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312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int INFint = 2147483647;
const long long INF = 9223372036854775807ll;
const long long MOD = 1000000007ll;
int main() {
ios_base::sync_with_stdio(0);
long long n, k;
cin >> n >> k;
if (n * (n - 1) / 2 <= k) {
cout << "no solution" << endl;
return 0;
}
for (int i = 0; i < n; i++) {
cout << 0 << ' ' << i << endl;
}
fprintf(stderr, "\nTIME = %lf\n", 1.0 * clock() / CLOCKS_PER_SEC);
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long int t, i, j, n, k, mx;
mx = 1000000000LL;
cin >> n >> k;
t = (n * (n - 1)) / 2;
if (t <= k) {
cout << "no solution";
} else {
for (i = 0; i < n - 1; i++) {
cout << "0 " << i << "\n";
}
cout << mx << " " << mx;
}
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
long n, k;
cin >> n >> k;
if (k >= n * (n - 1) / 2) {
cout << "no solution";
return 0;
}
for (long i = 0; i < n; i++) cout << "0 " << i << endl;
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class C {
BufferedReader reader;
StringTokenizer tokenizer;
PrintWriter out;
public void solve() throws IOException {
int N = nextInt();
int K = nextInt();
int max = (N-1)*N/2;
if( max <= K){
out.println("no solution");
return;
}
int max_num = 100000000;
out.println("0 0");
for(int i = 1; i < N; i++){
out.print(i + " ");
out.print( max_num );
max_num -= (N+1);
out.println();
}
}
/**
* @param args
*/
public static void main(String[] args) {
new C().run();
}
public void run() {
try {
reader = new BufferedReader(new InputStreamReader(System.in));
tokenizer = null;
out = new PrintWriter(System.out);
solve();
reader.close();
out.close();
} catch (Exception e) {
e.printStackTrace();
System.exit(1);
}
}
int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
String nextToken() throws IOException {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
tokenizer = new StringTokenizer(reader.readLine());
}
return tokenizer.nextToken();
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
int main() {
int n, k;
scanf("%d%d", &n, &k);
if (k >= n * (n - 1) / 2) {
printf("no solution\n");
return 0;
}
for (int i = 0; i < n; ++i) printf("0 %d\n", i);
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
if (n * (n - 1) / 2 <= k) {
cout << "no solution";
} else {
for (int i = 0; i < n; i++) cout << "0 " << i << "\n";
}
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.io.*;
import java.util.*;
import static java.lang.Math.*;
import static java.util.Arrays.*;
public class ClosestPair {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
//READ----------------------------------------------------
int n = sc.nextInt(), k = sc.nextInt();
if((n*(n-1))/2<=k)
System.out.println("no solution");
else{
StringBuilder sb = new StringBuilder();
for (int i = 0; i <= n-1; i++)
{
sb.append("0 "+i+"\n");
}
System.out.print(sb);
}
//CLOSE----------------------------------------------------
sc.close();
}
} | JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | n, k = map(int, raw_input().split())
n -= 1
if k >= n * (n + 1) / 2:
print "no solution"
else:
for i in range(n+1):
print 0, i | PYTHON |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int n, k, p1, p2;
int main() {
cin >> n >> k;
int tot = 0;
for (int i = 1; i <= n; ++i)
for (int j = i + 1; j <= n; ++j) ++tot;
if (tot > k) {
for (int i = 0; i < n; ++i) {
cout << 0 << ' ' << i << endl;
}
} else
cout << "no solution" << endl;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
if (k >= (n * (n - 1)) / 2) {
cout << "no solution";
return 0;
}
for (int i = 0; i < n; i++) {
cout << 0 << ' ' << i << endl;
}
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | n,k = [int(x) for x in raw_input().split()]
if k >= (n-1)*(n)/2:
print("no solution")
else:
while n:
print(1),n
n -= 1
| PYTHON |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 |
import java.io.InputStreamReader;
import java.io.PrintWriter;
import static java.lang.Math.*;
import java.util.ArrayList;
import java.util.Scanner;
/*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
/**
*
* @author pttrung
*/
public class C {
public static long Mod = 1000000009;
public static void main(String[] args) {
Scanner in = new Scanner(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(System.out);
int n = in.nextInt();
int k = in.nextInt();
long total = (n - 1)*(n)/2;
// System.out.println(total);
if (total <= k) {
out.println("no solution");
} else {
/**
* The solution will be in the line y = x
*/
out.println(0 + " " + 0);
long x = 0, y = 0;
for (int i = 1; i < n; i++) {
x += 0;
y += 1;
out.println(x + " " + y);
}
}
out.close();
}
static void check(Point a, Point b, ArrayList<Point> p, Point[] rec, int index) {
for (int i = 0; i < 4; i++) {
int m = (i + index) % 4;
int j = (i + 1 + index) % 4;
Point k = intersect(minus(b, a), minus(rec[m], rec[j]), minus(rec[m], a));
if (k.x >= 0 && k.x <= 1 && k.y >= 0 && k.y <= 1) {
Point val = new Point(k.x * minus(b, a).x, k.x * minus(b, a).y);
p.add(add(val, a));
// System.out.println(a + " " + b + " " + rec[i] + " " + rec[j]);
// System.out.println(add(val, a));
}
}
}
static Point intersect(Point a, Point b, Point c) {
double D = cross(a, b);
if (D != 0) {
return new Point(cross(c, b) / D, cross(a, c) / D);
}
return null;
}
static Point convert(Point a, double angle) {
double x = a.x * cos(angle) - a.y * sin(angle);
double y = a.x * sin(angle) + a.y * cos(angle);
return new Point(x, y);
}
static Point minus(Point a, Point b) {
return new Point(a.x - b.x, a.y - b.y);
}
static Point add(Point a, Point b) {
return new Point(a.x + b.x, a.y + b.y);
}
static double cross(Point a, Point b) {
return a.x * b.y - a.y * b.x;
}
static class Point {
double x, y;
Point(double x, double y) {
this.x = x;
this.y = y;
}
@Override
public String toString() {
return "Point: " + x + " " + y;
}
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | [n,k]=[int(i) for i in raw_input().split()]
if (n*(n-1)/2)<=k:
print "no solution"
else:
x=0
y=0
j=0
for i in range(n):
print 0,i
## print x,y
## j+=1
## if j==n:
## break
## print x+1,y
## j+=1
## if j==n:
## break
## print x+1,y+1
## j+=1
## if j==n:
## break
## print x,y+1
## j+=1
## if j==n:
## break
## x+=1
## y+=1
##epsilon=0.000000001
##[a,b,c,d]=[float(i) for i in raw_input().split()]
##j=0
##p=(a/b)
##q=((b-a)/b)*((d-c)/d)
##p1=p
##p2=p1+q*p
##i=2
##while p2-p1>epsilon:
## p1=p2
## p2=p2+(q**i)*p
## i+=1
##print p2
####n=int(raw_input())
####for i in range(n):
#### j=raw_input()
#### ##print j[-1:-5:-1]
#### if (j[-1:-6:-1]==".alal" and j[0:5]!="miao."):
#### print "Freda's"
#### elif (j[0:5]=="miao." and j[-1:-6:-1]!=".alal" ):
#### print "Rainbow's"
#### else:
#### print "OMG>.< I don't know!"
####
####
| PYTHON |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class B {
public static void main(String[] args) throws Exception {
Scanner sc = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
int n = sc.nextInt(), k = sc.nextInt();
if(1l*n*(n-1)/2 <= k) out.println("no solution");
else
for (int i = 0; i < n; i++) {
out.println("0 " + i);
}
out.flush();
out.close();
}
static class Scanner {
BufferedReader bf;
StringTokenizer st;
public Scanner(InputStream i) {
bf = new BufferedReader(new InputStreamReader(i));
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(bf.readLine());
return st.nextToken();
}
public int nextInt() throws NumberFormatException, IOException {
return Integer.parseInt(next());
}
public double nextDouble() throws IOException {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if (x.charAt(0) == '-') {
neg = true;
start++;
}
for (int i = start; i < x.length(); i++)
if (x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
} else {
sb.append(x.charAt(i));
if (dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public long nextLong() throws NumberFormatException, IOException {
return Long.parseLong(next());
}
}
} | JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
import java.math.BigInteger;
import java.math.RoundingMode;
public class a {
public static int vertices=0;
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n=s.nextInt();
int k=s.nextInt();
if(k>=(n*n-n)/2)
System.out.println("no solution");
else{
for(int i=0;i<n;i++)
System.out.println("1"+" "+(3*i));
}
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | //package Ladder_C;
/**
* Created by CompuShop on 7/25/2017 at 3:13 PM.
*/
import java.util.*;
public class C_312 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt(),k=in.nextInt();
if(k >= (n * (n - 1)) / 2)
{
System.out.println("no solution"); return;
}
for (int i = 0; i < n; i++) {
System.out.println(0+" "+i);
}
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
char a[1000];
void solve() {
int n, k;
while (scanf("%d%d", &n, &k) != EOF) {
if (((1 + n - 1) * (n - 1) / 2) <= k) {
printf("no solution\n");
continue;
} else {
int temp = 0;
for (int i = 1; i <= n; i++) {
printf("0 %d\n", i);
}
}
}
}
int main() {
solve();
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | n, k = map(int, input().split())
if (k >= n * (n - 1) // 2):
print("no solution")
else:
for i in range(n):
print(0, i)
| PYTHON3 |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
Scanner in = new Scanner(inputStream);
PrintWriter out = new PrintWriter(outputStream);
CTheClosestPair solver = new CTheClosestPair();
solver.solve(1, in, out);
out.close();
}
static class CTheClosestPair {
public void solve(int testNumber, Scanner sc, PrintWriter pw) {
int n = sc.nextInt();
int k = sc.nextInt();
int num = n * (n - 1) / 2;
if (k < num) {
int i = 0;
int j = 0;
boolean f = false;
while (n-- > 0) {
pw.println(i + " " + j);
j++;
}
} else pw.println("no solution");
}
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(FileReader r) {
br = new BufferedReader(r);
}
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
string c;
int main() {
int i, j, k, n, m, l;
scanf("%d%d", &n, &k);
if (n * (n - 1) / 2 <= k) {
puts("no solution");
return 0;
}
for (i = 1; i <= n; i++) printf("%d %d\n", 0, i);
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
struct node {
int x;
int y;
};
node point[2000 + 10];
int n;
int sum;
void solve() {
for (int i = 2; i <= n; ++i) {
point[i].y = point[i - 1].y + 1;
}
for (int i = 1; i <= n; ++i)
for (int j = i + 1; j <= n; ++j) ++sum;
}
int main() {
int k;
int tem;
scanf("%d%d", &n, &k);
solve();
if (k >= sum)
puts("no solution");
else {
for (int i = 1; i <= n; ++i) {
printf("%d %d\n", point[i].x, point[i].y);
}
}
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long N = 1e6 + 7;
long long n, k;
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cin >> n >> k;
long long mx = n * (n - 1) / 2;
if (k >= mx) return cout << "no solution", 0;
long long y = 0;
for (long long i = 1; i < n; i++, y += 2) cout << 0 << " " << y << "\n";
cout << 0 << " " << y - 1;
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.util.Scanner;
public class r185d2c {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
int k = scan.nextInt();
if((n*n-n)/2 <= k){
System.out.println("no solution");
} else {
for(int i=0; i < n; i++){
System.out.println("0 " + i);
}
}
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
inline long long inp() {
long long n = 0, s = 1;
char p = getchar();
if (p == '-') s = -1;
while ((p < '0' || p > '9') && p != EOF && p != '-') p = getchar();
if (p == '-') s = -1, p = getchar();
while (p >= '0' && p <= '9') {
n = (n << 3) + (n << 1) + (p - '0');
p = getchar();
};
return n * s;
}
long long power(long long a, long long b) {
long long r = 1, x = a;
if (a == 0) return 0;
while (b) {
if (b & 1) r = (r * x) % 1000000007;
x = (x * x) % 1000000007;
b >>= 1;
}
return r % 1000000007;
}
int gcd(int a, int b) {
if (b == 0) return a;
gcd(b, a % b);
}
long long distance(long long a, long long b, long long c, long long d) {
double ans;
ans = (a - c) * (a - c) + (b - d) * (b - d);
ans = sqrt(ans);
return ceil(ans);
}
int main() {
long long i, j, k, t, r, w, x, y, z, ans = 0, n = 0, m, d, sum, test, c = 0;
n = inp();
k = inp();
if (((n * (n - 1)) / 2) <= k) {
printf("no solution\n");
return 0;
}
for (i = 0; i < n; i++) cout << "0 " << i << endl;
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 |
import java.io.BufferedReader;
import java.io.Closeable;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class TheClosestPair implements Closeable {
private InputReader in = new InputReader(System.in);
private PrintWriter out = new PrintWriter(System.out);
public void solve() {
int n = in.ni(), k = in.ni();
if (n * (n - 1) <= 2 * k) {
out.println("no solution");
return;
}
for (int i = 0; i < n; i++) {
out.println(0 + " " + i);
}
}
@Override
public void close() throws IOException {
in.close();
out.close();
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int ni() {
return Integer.parseInt(next());
}
public long nl() {
return Long.parseLong(next());
}
public void close() throws IOException {
reader.close();
}
}
public static void main(String[] args) throws IOException {
try (TheClosestPair instance = new TheClosestPair()) {
instance.solve();
}
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
int main() {
int n, k;
scanf("%d%d", &n, &k);
if ((n * (n - 1) / 2) <= k) {
printf("no solution\n");
} else {
int i;
for (i = 0; i < n; i++) {
printf("0 %d\n", i);
}
}
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
template <class T>
inline T Max(T a, T b) {
if (a > b)
return a;
else
return b;
}
template <class T>
inline T Min(T a, T b) {
if (a < b)
return a;
else
return b;
}
template <class T>
inline T gcd(T a, T b) {
if (a < 0) return gcd(-a, b);
if (b < 0) return gcd(a, -b);
return (b == 0) ? a : gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
if (a < 0) return lcm(-a, b);
if (b < 0) return lcm(a, -b);
return a * (b / gcd(a, b));
}
template <class T>
inline T TripleMax(T a, T b, T c) {
return Max(Max(a, b), c);
}
template <class T>
inline T TripleMin(T a, T b, T c) {
return Min(Min(a, b), c);
}
const long long llinfinity = 9223372036854775807LL;
const long long llminusinfinity = -9223372036854775808LL;
const int intinfinity = 2147483647;
const int intminusinfinity = -2147483648;
int n, k;
int main(int argc, const char* argv[]) {
scanf("%d%d", &n, &k);
if (k >= (n * (n - 1) / 2)) {
puts("no solution");
} else {
for (int i = 0; i < n; ++i) {
printf("%d %d\n", 0, i);
}
}
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
int x[2010], y[2010];
using namespace std;
int main() {
int n, k;
cin >> n >> k;
for (int i = 0; i < n; ++i) x[i] = 0, y[i] = i;
int c = 0, tot = 0;
int d = (1 << 29);
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
tot++;
int t = (x[j] - x[i]) * (x[j] - x[i]) + (y[j] - y[i]) * (y[j] - y[i]);
if ((x[j] - x[i]) * (x[j] - x[i]) >= d) break;
if (d > t) d = t;
}
}
if (tot <= k)
cout << "no solution" << endl;
else
for (int i = 0; i < n; ++i) cout << x[i] << " " << y[i] << endl;
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | if __name__ == '__main__':
n, k = [int(x) for x in raw_input().rstrip().split()]
if n*(n-1) / 2 <= k:
print 'no solution'
else:
for i in range(n):
print '0 {}'.format(i) | PYTHON |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
int main() {
int cn = 0, i, n, k, p;
scanf("%d%d", &n, &k);
if (k >= (n * (n - 1)) / 2) {
printf("no solution\n");
return (0);
}
p = n;
while (p--) {
printf("0 %d\n", cn);
cn++;
}
return (0);
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int n, k;
int main() {
cin >> n >> k;
int temp = n * (n - 1) / 2;
if (temp <= k) {
cout << "no solution\n";
return 0;
}
for (int i = 0; i < n; i++) {
cout << 0 << " " << i << endl;
}
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int MAX = 2001;
int N, K;
int main() {
scanf("%d%d", &N, &K);
if (K >= N * (N - 1) / 2) {
puts("no solution");
} else {
for (int i = 0; i < N; i++) {
printf("0 %d\n", i);
}
}
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import static java.util.Arrays.deepToString;
import java.io.*;
import java.math.*;
import java.util.*;
public class Main {
static void solve() {
int n = nextInt();
int k = nextInt();
int[] x = new int[n];
int[] y = new int[n];
if (n * (n - 1) <= 2 * k) {
System.out.println("no solution");
} else {
for (int i = 0; i < n; i++) {
y[i] = i;
}
for (int i = 0; i < n; i++) {
System.out.println(x[i] + " " + y[i]);
}
}
}
public static void main(String[] args) throws Exception {
reader = new BufferedReader(new InputStreamReader(System.in));
writer = new PrintWriter(System.out);
setTime();
solve();
printTime();
printMemory();
writer.close();
}
static BufferedReader reader;
static PrintWriter writer;
static StringTokenizer tok = new StringTokenizer("");
static long systemTime;
static void debug(Object... o) {
System.err.println(deepToString(o));
}
static void setTime() {
systemTime = System.currentTimeMillis();
}
static void printTime() {
System.err.println("Time consumed: "
+ (System.currentTimeMillis() - systemTime));
}
static void printMemory() {
System.err.println("Memory consumed: "
+ (Runtime.getRuntime().totalMemory() - Runtime.getRuntime()
.freeMemory()) / 1000 + "kb");
}
static String next() {
while (!tok.hasMoreTokens()) {
String w = null;
try {
w = reader.readLine();
} catch (Exception e) {
e.printStackTrace();
}
if (w == null)
return null;
tok = new StringTokenizer(w);
}
return tok.nextToken();
}
static int nextInt() {
return Integer.parseInt(next());
}
static long nextLong() {
return Long.parseLong(next());
}
static double nextDouble() {
return Double.parseDouble(next());
}
static BigInteger nextBigInteger() {
return new BigInteger(next());
}
} | JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
#pragma warning(disable : 4996)
struct point {
int x, y;
};
int n, k;
int dx[2010], dy[2010];
int tot;
int sqr(int x) { return x * x; };
int imin(int a, int b) {
if (a < b) return a;
return b;
};
void algo() {
int d = 2100000000;
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++) {
tot++;
if (dx[j] - dx[i] >= d) break;
d = imin(d, sqr(dx[i] - dx[j]) + sqr(dy[i] - dy[j]));
};
};
int main() {
cin >> n >> k;
for (int i = 0; i < n; i++) {
dx[i] = 0;
dy[i] = i;
};
algo();
if (tot <= k)
cout << "no solution";
else {
for (int i = 0; i < n; i++) cout << dx[i] << " " << dy[i] << endl;
};
return 0;
};
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | n, k = map(int, input().split())
tot = 0
for i in range(n):
for j in range(i + 1, n):
tot += 1
if tot <= k:
print('no solution')
else:
for i in range(n):
print(1, i)
| PYTHON3 |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | n,k = map(int ,input().split())
if((n&1)==0) :
time = (n//2)*(n-1)
else :
time = ((n-1)//2)*n
if(k>=time):
print("no solution")
else :
for i in range(n):
print("0 "+str(i)) | PYTHON3 |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | /**
* Created with IntelliJ IDEA.
* User: Venky
*/
import java.io.*;
import java.util.StringTokenizer;
public class Main {
static void solve() throws IOException {
int n = nextInt();
int k = nextInt();
if( ((n-1)*n)/2 <= k)
{
out.println("no solution");
return;
}
for(int i=0;i<n;i++)
{
out.println(i + " " + (i*3000));
}
}
static BufferedReader br;
static StringTokenizer st;
static PrintWriter out;
public static void main(String[] args) throws IOException {
InputStream input = System.in;
PrintStream output = System.out;
br = new BufferedReader(new InputStreamReader(input));
out = new PrintWriter(output);
solve();
out.close();
}
static long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
static double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
static int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
static String nextToken() throws IOException {
while (st == null || !st.hasMoreTokens()) {
String line = br.readLine();
if (line == null) {
return null;
}
st = new StringTokenizer(line);
}
return st.nextToken();
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, i, k;
int tot;
cin >> n >> k;
tot = (n * (n - 1)) / 2;
if (tot <= k) {
cout << "no solution\n";
return 0;
}
for (i = 0; i < n; i++) {
cout << i << " " << i * 100000 << "\n";
}
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n, k;
cin >> n >> k;
long long tot;
tot = (n * (n - 1)) / 2;
if (tot <= k) {
cout << "no solution";
return 0;
}
for (int i = 0; i < n; i++) cout << 0 << " " << i << endl;
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import sys
import itertools
def solve(n, k):
comb = 0
for x in itertools.combinations([0]*n, 2):
comb += 1
if comb<=k:
return ["no solution"]
res = []
#base = 2**n
for i in xrange(n):
#res.append(str(base/2**(i)) + " -" + str(base*2**(i)))
res.append('0 -'+str(i+1))
return res
n, k = map(int, sys.stdin.readline().split())
res = solve(n, k)
for string in res:
print string
| PYTHON |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
int l = n * (n - 1);
l /= 2;
if (k >= l) {
cout << "no solution\n";
return 0;
}
pair<int, int> p;
vector<pair<int, int> > v;
int a = 0, b = 0;
for (int i = 0; i < n; i++) {
p = make_pair(b, a);
v.push_back(p);
a++;
}
for (int i = 0; i < n; i++) {
cout << v[i].first << " " << v[i].second << endl;
}
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
public static void main(String [] args ) {
try{
String str;
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedOutputStream bos = new BufferedOutputStream(System.out);
String eol = System.getProperty("line.separator");
byte [] eolb = eol.getBytes();
byte[] spaceb= " ".getBytes();
str = br.readLine();
int blank = str.indexOf( " ");
int n = Integer.parseInt(str.substring(0,blank));
int m = Integer.parseInt(str.substring(blank+1));
int tot = ( n * (n-1) ) / 2 ;
if(m>= tot) {
bos.write("no solution".getBytes());
bos.write(eolb);
} else {
for(int i = 0 ; i < n ; i++) {
bos.write("0".getBytes());
bos.write(spaceb);
bos.write(new Integer(i).toString().getBytes());
bos.write(eolb);
}
}
bos.flush();
} catch(IOException ioe) {
ioe.printStackTrace();
}
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.util.Scanner;
public class C185 {
public void run() {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int k = in.nextInt();
if (n * (n - 1) <= 2 * k) {
System.out.println("no solution");
System.exit(0);
}
int x[] = new int[n + 100];
int y[] = new int[n + 100];
x[0] = -1000000000;
y[0] = -1000000000;
int d = 10000;
for (int i = 1; i < n + 5; i++) {
x[i] = x[0] + i;
y[i] = y[0] + i * d;
}
int arr[] = new int[n + 10];
arr[0] = 1;
arr[1] = 1;
for (int i = 2; i < n + 10; i++)
arr[i] = arr[i - 1] + i - 1;
// for (int i = 0; i < n + 5; i++)
// System.out.println(arr[i]);
for (int i = 1; i < n + 5; i++) {
//x[i] += arr[i];
y[i] += arr[i];
}
for (int i = 0; i < n; i++)
System.out.println(x[i] + " " + y[i]);
}
public static void main(String[] args) {
// TODO Auto-generated method stub
new C185().run();
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.io.*;
import java.util.*;
public class C
{
String line;
StringTokenizer inputParser;
BufferedReader is;
FileInputStream fstream;
DataInputStream in;
String FInput="";
void openInput(String file)
{
if(file==null)is = new BufferedReader(new InputStreamReader(System.in));//stdin
else
{
try{
fstream = new FileInputStream(file);
in = new DataInputStream(fstream);
is = new BufferedReader(new InputStreamReader(in));
}catch(Exception e)
{
System.err.println(e);
}
}
}
void readNextLine()
{
try {
line = is.readLine();
inputParser = new StringTokenizer(line, " ");
//System.err.println("Input: " + line);
} catch (IOException e) {
System.err.println("Unexpected IO ERROR: " + e);
}
}
int NextInt()
{
String n = inputParser.nextToken();
int val = Integer.parseInt(n);
//System.out.println("I read this number: " + val);
return val;
}
String NextString()
{
String n = inputParser.nextToken();
return n;
}
void closeInput()
{
try {
is.close();
} catch (IOException e) {
System.err.println("Unexpected IO ERROR: " + e);
}
}
public void readFInput()
{
for(;;)
{
try
{
readNextLine();
FInput+=line+" ";
}
catch(Exception e)
{
break;
}
}
inputParser = new StringTokenizer(FInput, " ");
}
long NextLong()
{
String n = inputParser.nextToken();
long val = Long.parseLong(n);
return val;
}
public static void main(String [] argv)
{
String filePath=null;
if(argv.length>0)filePath=argv[0];
new C(filePath);
}
boolean [] [] p;
public C(String inputFile)
{
openInput(inputFile);
readNextLine();
int n=NextInt(), k=NextInt();
if(n*n-n>k*2)
{
for(int i=0; i<n; i++)
System.out.println(47+" "+i);
}
else
System.out.println("no solution");
closeInput();
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
#pragma comment(linker, "/STACK:256777216")
using namespace std;
int main() {
int n, k;
cin >> n >> k;
if (2 * k >= n * (n - 1)) {
puts("no solution");
return 0;
}
for (int i = 1; i <= n; i++) {
cout << 0 << " " << i << endl;
}
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 |
import java.io.*;
import java.util.*;
import java.math.*;
import static java.lang.Math.*;
import static java.lang.Integer.*;
// WA on 5
public class C185 {
int INF = Integer.MAX_VALUE / 100;
static Scanner sc = null;
static BufferedReader br = null;
static PrintStream out = null;
static BufferedWriter bw = null;
int N = 0;
public void solve() throws Exception{
int n = sc.nextInt();
int k = sc.nextInt();
int sum = 0;
for(int i = n-1; i >= 0; i--){
sum += i;
}
if(sum <= k){
out.println("no solution");
return;
}
for(int i = 0; i < n; i++){
out.println("0 " + i);
}
}
public int[] readIntArray(int n) {
int[] ret = new int[n];
for (int i = 0; i < n; i++) {
ret[i] = sc.nextInt();
}
return ret;
}
/**
* @param args
*/
public static void main(String[] args) throws Exception{
File file = new File("input.txt");
if(file.exists()){
System.setIn(new BufferedInputStream(new FileInputStream("input.txt")));
}
out = System.out;
bw = new BufferedWriter(new PrintWriter(out));
sc = new Scanner(System.in);
//br = new BufferedReader(new InputStreamReader(System.in));
C185 t = new C185();
t.solve();
bw.close();
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.io.*;
import java.util.*;
public class SolutionC {
BufferedReader in;
StringTokenizer str;
PrintWriter out;
String SK;
String next() throws IOException {
while ((str == null) || (!str.hasMoreTokens())) {
SK = in.readLine();
if (SK == null)
return null;
str = new StringTokenizer(SK);
}
return str.nextToken();
}
int nextInt() throws IOException {
return Integer.parseInt(next());
}
double nextDouble() throws IOException {
return Double.parseDouble(next());
}
long nextLong() throws IOException {
return Long.parseLong(next());
}
char[] charArray() throws IOException{
return next().toCharArray();
}
public static void main(String[] args) throws IOException {
new SolutionC().run();
}
void run() throws IOException {
in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
//in = new BufferedReader(new FileReader("input.txt"));
//out = new PrintWriter("output.txt");
solve();
out.close();
}
void solve() throws IOException {
long n = nextLong();
long k = nextLong();
if(n*(n-1)/2<=k){
out.println("no solution");
}
else{
for (int i = 0; i < n; i++) {
out.println(0+" "+i);
}
}
}
}
class P{
int x;
int y;
} | JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
int cnt = (n * (n - 1)) / 2;
if (cnt <= k)
cout << "no solution";
else {
for (int i = 1; i <= n; i++) {
cout << 0 << ' ' << i << '\n';
}
}
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | ar = raw_input().split(" ")
n = int(ar[0])
k = int(ar[1])
tot = (n*(n-1))/2
if tot <= k:
print "no solution"
else:
t = n
a = 0
b = 0
while t:
print a, b
a = a+1
b = b+n
t-=1 | PYTHON |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class C312 {
static StringTokenizer st;
static BufferedReader in;
static PrintWriter pw;
public static void main(String[] args) throws IOException {
in = new BufferedReader(new InputStreamReader(System.in));
pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
int n = nextInt();
int k = nextInt();
if (n*(n-1)/2 <= k)
System.out.println("no solution");
else {
for (int i = 1; i <= n; i++) {
System.out.println(0+" "+i);
}
}
pw.close();
}
private static int nextInt() throws IOException{
return Integer.parseInt(next());
}
private static long nextLong() throws IOException{
return Long.parseLong(next());
}
private static double nextDouble() throws IOException{
return Double.parseDouble(next());
}
private static String next() throws IOException {
while (st == null || !st.hasMoreTokens()) {
st = new StringTokenizer(in.readLine());
}
return st.nextToken();
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int t, n, k, ax, bx, tot;
cin >> n >> k;
tot = (n * (n + 1)) / 2 - n;
if (tot <= k)
cout << "no solution" << endl;
else {
for (int i = 1; i <= n; i++) {
cout << 0 << " " << i << endl;
}
}
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
unsigned long long mod = 1000000007;
int main() {
int nop, k;
cin >> nop >> k;
int max_possi = nop * (nop - 1);
max_possi /= 2;
if (k >= max_possi)
cout << "no solution"
<< "\n";
else {
int x = 0, y = 0;
while (nop--) cout << x << " " << y++ << "\n";
}
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long distance(long long x1, long long y1, long long x2, long long y2) {
return sqrt(pow(x2 - x1, 2) + pow(y2 - y1, 2));
}
int main() {
long long n, k, d, x, y, tot = 0;
cin >> n >> k;
for (int i = 1; i <= n; i++) {
for (int j = i + 1; j <= n; j++) {
tot++;
}
}
if (tot <= k)
cout << "no solution" << endl;
else {
for (int i = 0; i < n; i++) {
cout << 0 << " " << i + 12345 << endl;
}
}
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
if (n * (n - 1) <= 2 * k)
cout << "no solution" << endl;
else
for (int y = 0; y < n; y++) {
cout << 0 << " " << y << endl;
}
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.util.*;
import java.io.*;
public class third
{
public static void main(String args[])
{
int n,k;
Scanner sc = new Scanner(System.in);
n=sc.nextInt();
k=sc.nextInt();
if(k>=((n*(n-1))/2))
{
System.out.println("no solution");
}
else
{
int f=0,t=0;
while(t<n)
{
System.out.println(1+" "+t);
t++;
}
}
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.io.IOException;
import java.io.InputStream;
import java.util.InputMismatchException;
public class CF {
public static void main(String[] args) {
FasterScanner sc = new FasterScanner();
int N =sc.nextInt();
int K =sc.nextInt();
if(((N*(N-1))>>1)<=K){
System.out.println("no solution");
return;
}
for(int a=0;a<N;a++){
System.out.println(a+" "+(a*200000));
}
}
static class FasterScanner{
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
public FasterScanner(){
stream = System.in;
//stream = new FileInputStream(new File("dec.in"));
}
int read(){
if(numChars==-1)
throw new InputMismatchException();
if(curChar>=numChars){
curChar = 0;
try{
numChars = stream.read(buf);
} catch (IOException e){
throw new InputMismatchException();
}
if(numChars <= 0)
return -1;
}
return buf[curChar++];
}
boolean isSpaceChar(int c){
return c==' '||c=='\n'||c=='\r'||c=='\t'||c==-1;
}
boolean isEndline(int c){
return c=='\n'||c=='\r'||c==-1;
}
int nextInt(){
return Integer.parseInt(next());
}
long nextLong(){
return Long.parseLong(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
String next(){
int c = read();
while(isSpaceChar(c))
c=read();
StringBuilder res = new StringBuilder();
do{
res.appendCodePoint(c);
c=read();
} while(!isSpaceChar(c));
return res.toString();
}
String nextLine(){
int c = read();
while(isEndline(c))
c=read();
StringBuilder res = new StringBuilder();
do{
res.appendCodePoint(c);
c = read();
}while(!isEndline(c));
return res.toString();
}
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long n, k;
int main() {
ios_base::sync_with_stdio(0);
cin >> n >> k;
if (n * (n - 1) / 2 <= k)
cout << "no solution" << endl;
else
for (int i = 0; i < n; i++) cout << 0 << " " << i << endl;
cin.get();
cin.get();
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
int x = n - 1, tot = 0;
tot = (x * (x + 1)) / 2;
if (tot <= k) return cout << "no solution" << endl, 0;
for (int i = 0; i < n; i++) cout << 1 << " " << i << endl;
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | n, k = map(int, raw_input().split())
if n*(n-1)/2 <= k:
print "no solution"
else:
for i in range(0, n):
print 0, i
| PYTHON |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | f = lambda n: n*(n-1)/2
n,k = map(int, raw_input().split())
if k>=f(n):
print 'no solution'
else:
for i in xrange(n):
print 0,i | PYTHON |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.Scanner;
public class Main implements Runnable {
StreamTokenizer ST;
PrintWriter out;
BufferedReader br;
Scanner in;
public static void main(String[] args) throws IOException {
new Thread(new Main()).start();
}
@Override
public void run() {
try {
out = new PrintWriter(new BufferedOutputStream(System.out));
in = new Scanner(System.in);
solve();
out.close();
} catch (IOException e) {
throw new IllegalStateException(e);
}
}
public void solve() throws IOException {
int n = in.nextInt();
int k = in.nextInt();
if (k >= (n * (n - 1)) / 2) {
out.println("no solution");
} else {
int z = (int) 5e4;
out.println("0 0");
int t = 0;
for (int i = 1; i < n; ++i) {
t += (z - (i + 1));
out.println("0 " + t);
}
}
}
} | JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.util.*;
import java.io.*;
public class C312 {
class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream));
tokenizer = null;
}
public InputReader() throws FileNotFoundException {
reader = new BufferedReader(new FileReader("d:/input.txt"));
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public double nextDouble(){
return Double.parseDouble(next());
}
public long nextLong(){
return Long.parseLong(next());
}
}
public void run(){
InputReader reader = new InputReader(System.in);
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
int n = reader.nextInt(), k = reader.nextInt();
int max = (n-1)*n/2;
if(k >= max){
out.println("no solution");
}else{
for(int i = 0 ; i < n ; i ++){
out.println(0+" "+i);
}
}
out.flush();
}
public static void main(String[] args) {
new C312().run();
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.math.*;
import java.util.*;
public class CF {
public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int k = in.nextInt();
if(((n * (n-1))/2) > k)
{
int m = 0;
int j = 0;
while(m++ < n)
{
System.out.println("0 " + j++);
}
}
else
{
System.out.print("no solution");
return;
}
}
public static int factorial(int a)
{
if(a == 1)
return 1;
return a * factorial(a-1);
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, tot;
scanf("%d%d", &n, &tot);
if (n * (n - 1) / 2 <= tot) {
printf("no solution");
return 0;
}
for (int i = 0; i < n; i++) printf("%d %d\n", 0, i);
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.util.Scanner;
public class ProblemC {
public void solve() {
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
int k = scan.nextInt();
if ((n * (n - 1)) / 2 <= k) {
System.out.println("no solution");
return;
}
for (int i = 1; i <= n; i++) {
System.out.println("0 " + i);
}
}
public static void main(String[] args) {
ProblemC c = new ProblemC();
c.solve();
}
} | JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.util.*;
import java.io.*;
import java.math.*;
public class Main
{
static class Reader
{
private InputStream mIs;private byte[] buf = new byte[1024];private int curChar,numChars;public Reader() { this(System.in); }public Reader(InputStream is) { mIs = is;}
public int read() {if (numChars == -1) throw new InputMismatchException();if (curChar >= numChars) {curChar = 0;try { numChars = mIs.read(buf);} catch (IOException e) { throw new InputMismatchException();}if (numChars <= 0) return -1; }return buf[curChar++];}
public String nextLine(){int c = read();while (isSpaceChar(c)) c = read();StringBuilder res = new StringBuilder();do {res.appendCodePoint(c);c = read();}while (!isEndOfLine(c));return res.toString() ;}
public String s(){int c = read();while (isSpaceChar(c)) c = read();StringBuilder res = new StringBuilder();do {res.appendCodePoint(c);c = read();}while (!isSpaceChar(c));return res.toString();}
public long l(){int c = read();while (isSpaceChar(c)) c = read();int sgn = 1;if (c == '-') { sgn = -1 ; c = read() ; }long res = 0; do{ if (c < '0' || c > '9') throw new InputMismatchException();res *= 10 ; res += c - '0' ; c = read();}while(!isSpaceChar(c));return res * sgn;}
public int i(){int c = read() ;while (isSpaceChar(c)) c = read();int sgn = 1;if (c == '-') { sgn = -1 ; c = read() ; }int res = 0;do{if (c < '0' || c > '9') throw new InputMismatchException();res *= 10 ; res += c - '0' ; c = read() ;}while(!isSpaceChar(c));return res * sgn;}
public double d() throws IOException {return Double.parseDouble(s()) ;}
public boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; }
public boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; }
}
///////////////////////////////////////////////////////////////////////////////////////////
// RRRRRRRRR AAA HHH HHH IIIIIIIIIIIII LLL //
// RR RRR AAAAA HHH HHH IIIIIIIIIII LLL //
// RR RRR AAAAAAA HHH HHH III LLL //
// RR RRR AAA AAA HHHHHHHHHHH III LLL //
// RRRRRR AAA AAA HHHHHHHHHHH III LLL //
// RR RRR AAAAAAAAAAAAA HHH HHH III LLL //
// RR RRR AAA AAA HHH HHH IIIIIIIIIII LLLLLLLLLLLL //
// RR RRR AAA AAA HHH HHH IIIIIIIIIIIII LLLLLLLLLLLL //
///////////////////////////////////////////////////////////////////////////////////////////
static int n;
static long store[];
public static void main(String[] args)throws IOException
{
PrintWriter out= new PrintWriter(System.out);
Reader sc=new Reader();
int n=sc.i();
int k=sc.i();
if(k>=(n*(n-1))/2)
out.println("no solution");
else
{
for(int i=0;i<n;i++)
out.println("0 "+i);
}
out.flush();
}
} | JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | # -*- coding: utf-8 -*-
n, k = (int(x) for x in raw_input().split())
m = (n - 1) * n / 2
if k >= m:
print 'no solution'
else:
for i in range(n):
print '{} {}'.format(0, i) | PYTHON |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #!/usr/bin/env python
import sys
n, k = map(int, sys.stdin.readline().strip().split())
mx = (n-1)*n / 2
if k >= mx:
print 'no solution'
else:
for i in xrange(n):
print 0, i
| PYTHON |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.util.*;
public class ClosestPair {
public static void main(String[] args)
{
Scanner s=new Scanner(System.in);
int n=s.nextInt();
int k=s.nextInt();
point[] p=new point[n];
int c=0;
for(int i=0;i<n;i++)
{ point pt = new point();
pt.x=0;
pt.y=c;
c++;
p[i]=pt;
}
Arrays.sort(p,point.ySort);
double d=Integer.MAX_VALUE;
int tot=0;
for(int i=0;i<n;i++)
for(int j=i+1;j<n;j++)
{
++tot;
if (p[j].x-p[i].x>=d)
break;
d=Math.min(d,distance(p[i],p[j]));
}
if(tot<=k)
System.out.println("no solution");
else
for(int i=0;i<n;i++)
{
System.out.println(p[i].x+" " + p[i].y);
}
}
public static double distance(point p1,point p2)
{
return (Math.sqrt(Math.pow(p1.x-p2.x, 2)+Math.pow(p1.y-p2.y, 2)));
}
static class point implements Comparable<point>
{
int x,y;
public point()
{
}
public point(int x, int y)
{
this.x=x;
this.y=y;
}
public int compareTo(point pt) {
return this.x-pt.x;
}
public static Comparator<point> ySort =new Comparator<point>()
{
public int compare(point p1,point p2) {
int x= p1.x-p2.x;
if(p1.x!=p2.x)
return x;
else
return p1.y-p2.y;
}
};
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | n,k = map(int, raw_input().split())
if n*(n-1)/2 <= k:
print "no solution"
else:
for i in range(n):
print 0,i
| PYTHON |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
int tot = (n) * (n - 1);
tot /= 2;
if (tot <= k)
cout << "no solution" << endl;
else {
for (int i = 0; i < n; ++i) cout << 1 << " " << i + 1 << endl;
}
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
double distance(pair<int, int> a, pair<int, int> b) {
return pow(a.first - b.first, 2.0) + pow(a.second - b.second, 2.0);
}
int gettot(vector<pair<int, int> > v) {
sort(v.begin(), v.end());
double d = 1e100;
int tot = 0;
for (int i = 0; i < v.size(); i++) {
for (int j = i + 1; j < v.size(); j++) {
++tot;
if (v[j].first - v[i].first >= d) break;
}
}
return tot;
}
int main() {
int n = 4;
int k = 3;
cin >> n >> k;
int count = 0;
int max = (n - 1) * n / 2;
if (k >= max) {
cout << "no solution" << endl;
return 0;
}
int ys = 0;
vector<pair<int, int> > v;
for (int i = 1; i <= n; i++) {
cout << 1 << " " << ys << endl;
ys += 10;
}
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.io.*;
import java.util.*;
public class main
{
public static void main(String[] args) throws Exception
{
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int k = in.nextInt();
int sum = (n * (n-1)) / 2;
if(sum <= k)
{
System.out.println("no solution");
}
else
{
for(int i = 1; i <= n; i++)
{
System.out.println(i + " " + (i*30000));
}
}
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | n,k=map(int,raw_input().split())
y=sum([x for x in range(1,n)])
if y>k:
p=1
for i in range(n):
print "0 "+str(p)
p+=1
else:
print "no solution" | PYTHON |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | n,k=map(int,input().split())
if n*(n-1)/2<=k:
print("no solution")
else:
for i in range(0,n):print(0,i)
# Made By Mostafa_Khaled | PYTHON3 |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
if ((n * (n - 1)) / 2 <= k) {
cout << "no solution\n";
} else {
for (int i = 0; i < n; i++) {
cout << "0 " << i << endl;
}
}
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | n,k = map(int, raw_input().split())
if k < n*(n-1)/2:
for i in xrange(n):
print 0, i
else:
print "no solution" | PYTHON |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int i, j, n;
long long k;
while (~scanf("%d%I64d", &n, &k)) {
long long t = (long long)n * (n - 1) / 2;
if (k >= t) {
printf("no solution\n");
continue;
}
int y = 1;
for (i = 0; i < n; i++) printf("%d %d\n", 0, y), y++;
}
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class Main
{
static BufferedReader reader;
static StringTokenizer tokenizer;
static PrintWriter writer;
static int nextInt() throws IOException
{
return Integer.parseInt(nextToken());
}
static long nextLong() throws IOException
{
return Long.parseLong(nextToken());
}
static double nextDouble() throws IOException
{
return Double.parseDouble(nextToken());
}
static boolean eof = false;
static String nextToken() throws IOException
{
while (tokenizer == null || !tokenizer.hasMoreTokens())
{
tokenizer = new StringTokenizer(reader.readLine());
}
return tokenizer.nextToken();
}
public static void main(String[] args) throws IOException
{
tokenizer = null;
reader = new BufferedReader(new InputStreamReader(System.in));
writer = new PrintWriter(System.out);
banana();
reader.close();
writer.close();
}
static int test(int a[], int b[])
{
int sum = 0;
for (int i = 0; i < a.length; ++i)
sum += a[i] * b[i];
return sum;
}
static void banana() throws IOException
{
int n = nextInt();
int k = nextInt();
if (k >= n * (n - 1) / 2)
System.out.println("no solution");
else
{
for (int i = 0; i < n; ++i)
{
System.out.println(i + " " +i * 2 * n) ;
}
}
}
} | JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int N = 105;
const int M = 505;
const int MOD = int(1e9) + 7;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-9;
const double PI = acos(-1.0);
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};
template <class T>
inline T Min(T a, T b) {
return a < b ? a : b;
}
template <class T>
inline T Max(T a, T b) {
return a > b ? a : b;
}
template <class T>
inline T Min(T a, T b, T c) {
return min(min(a, b), c);
}
template <class T>
inline T Max(T a, T b, T c) {
return max(max(a, b), c);
}
template <class T>
inline T sqr(T a) {
return a * a;
}
template <class T>
inline T cub(T a) {
return a * a * a;
}
template <class T>
inline T gcd(T a, T b) {
return b == 0 ? a : gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
return a * b / gcd(a, b);
}
int main() {
int a, b;
while (scanf("%d%d", &a, &b) != EOF) {
int i, tot = (a * (a - 1)) / 2;
if (tot <= b) {
printf("no solution\n");
} else {
for (i = 0; i < a; i++) {
printf("%d %d\n", 0, i);
}
}
}
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | n,k = map(int,raw_input().split())
if n == 2 :
print "no solution"
exit()
lim = n*(n-1)/2
if k>=lim:
print "no solution"
else:
for _ in xrange(n):
print "0 "+str(_)
| PYTHON |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | n, k = map(int, raw_input().split())
if k >= (n * (n - 1)) / 2:
print 'no solution'
exit()
for i in range(n):
print 0, i
| PYTHON |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
if (n * (n - 1) / 2 <= k)
cout << "no solution";
else
for (int i = 0; i < n; ++i) cout << "0 " << i << endl;
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
int main() {
int n, k, i;
scanf("%d %d", &n, &k);
if (k < n * (n - 1) / 2) {
for (i = 0; i < n; i++) printf("%d %d\n", 0, 3 * i);
} else
puts("no solution");
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.util.*;
import java.math.*;
import java.io.*;
public class Main
{
public static void main(String[] args) throws IOException
{
// BufferedReader c = new BufferedReader(new InputStreamReader(System.in));
Scanner c=new Scanner(System.in);
int n=c.nextInt(),k=c.nextInt();
int largestPossibleTime=(n*(n-1))/2;
if (k>=largestPossibleTime) System.out.println("no solution");
else
{
for (int i=0;i<n;i++) System.out.println(0+" "+i);
}
}
}
//must declare new classes here | JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.util.*;
import java.util.Map.Entry;
import java.io.*;
import java.awt.Point;
import java.math.BigInteger;
import static java.lang.Math.*;
public class Codeforces_Solution_C implements Runnable{
final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;
BufferedReader in;
PrintWriter out;
StringTokenizer tok = new StringTokenizer("");
void init() throws FileNotFoundException{
if (ONLINE_JUDGE){
in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
}else{
in = new BufferedReader(new FileReader("input.txt"));
out = new PrintWriter("output.txt");
}
}
String readString() throws IOException{
while(!tok.hasMoreTokens()){
try{
tok = new StringTokenizer(in.readLine());
}catch (Exception e){
return null;
}
}
return tok.nextToken();
}
int readInt() throws IOException{
return Integer.parseInt(readString());
}
long readLong() throws IOException{
return Long.parseLong(readString());
}
double readDouble() throws IOException{
return Double.parseDouble(readString());
}
public static void main(String[] args){
new Thread(null, new Codeforces_Solution_C(), "", 128 * (1L << 20)).start();
}
long timeBegin, timeEnd;
void time(){
timeEnd = System.currentTimeMillis();
System.err.println("Time = " + (timeEnd - timeBegin));
}
long memoryTotal, memoryFree;
void memory(){
memoryFree = Runtime.getRuntime().freeMemory();
System.err.println("Memory = " + ((memoryTotal - memoryFree) >> 10) + " KB");
}
void debug(Object... objects){
if (DEBUG){
for (Object o: objects){
System.err.println(o.toString());
}
}
}
public void run(){
try{
timeBegin = System.currentTimeMillis();
memoryTotal = Runtime.getRuntime().freeMemory();
init();
solve();
out.close();
time();
memory();
}catch (Exception e){
e.printStackTrace(System.err);
System.exit(-1);
}
}
boolean DEBUG = false;
void solve() throws IOException{
int n = readInt();
int k = readInt();
if (k>=(n*(n-1))/2) {
out.println("no solution");
return;
}
for (int i=0; i<n; i++) out.println(i+" "+i*n);
}
} | JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import sys
def main():
n, k = map(int, sys.stdin.readline().split())
if n * (n - 1) / 2 <= k:
print "no solution"
return
for i in xrange(n):
print 0, i
return
if __name__ == "__main__":
main() | PYTHON |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.io.IOException;
import java.util.Scanner;
/**
* Created with IntelliJ IDEA.
* User: Alexander Shchegolev
* Date: 26.05.13
* Time: 18:08
* To change this template use File | Settings | File Templates.
*/
public class Three {
public static void main(String[] args) throws IOException {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int k = scanner.nextInt();
if (n * (n - 1) / 2 <= k) {
System.out.println("no solution");
return;
}
for (int i = 0; i < n; i++) {
int x = i / 500000000;
int y = i % 500000000;
System.out.printf("%d %d\n", x, y);
}
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.util.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
Scanner in = new Scanner(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskC solver = new TaskC();
solver.solve(1, in, out);
out.close();
}
}
class TaskC {
public void solve(int testNumber, Scanner in, PrintWriter out) {
int n = in.nextInt();
int k = in.nextInt();
int sum = (n * (n - 1)) / 2;
if (sum <= k) {
out.println("no solution");
return;
}
for (int i = 0; i < n; i++) {
out.println("0 " + i);
}
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.util.*;
import java.io.*;
import java.math.*;
public class C {
public static Scanner scan = new Scanner(System.in);
public static boolean bg = false;
public static void main(String[] args) throws Exception {
long n1 = Integer.parseInt(scan.next());
long n2 = Integer.parseInt(scan.next());
long bound = n1*(n1-1)/2;
if (bg) System.out.println(bound);
if (n2>=bound){
System.out.println("no solution");
}
else {
for (int i=0;i<n1;i++){
System.out.println(0+" "+i);
}
}
}
}
| JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
long long k;
cin >> n >> k;
int ans = (n * (n - 1)) / 2;
if (k >= ans) {
cout << "no solution" << endl;
return 0;
} else {
for (int i = 0; i < n; i++) {
cout << 0 << ' ' << i << endl;
}
}
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | n, k = map(int, input().split())
up = (int)(n * (n - 1) / 2);
if k >= up:
print ("no solution")
else:
for i in range (0, n):
print (0, i)
| PYTHON3 |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int k, n;
int main() {
int i, j;
scanf("%d%d", &n, &k);
if (k >= (n * (n - 1) / 2))
puts("no solution");
else {
for (i = 1; i <= n; ++i) printf("0 %d\n", i);
}
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
string mirror = "AHIMOTUVWXY";
string letter = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const int mxn = 1e6 + 5;
const int mod = 1e9 + 7;
int main() {
long long n, m, i, j, k, x, y, t;
cin >> n >> k;
x = n * (n - 1) / 2;
if (x <= k) {
cout << "no solution" << endl;
} else {
for (i = 0; i < n; i++) cout << 0 << " " << i << endl;
}
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | /*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
import java.io.*;
import java.util.*;
/**
*
* @author N-AssassiN
*/
public class Main {
private static BufferedReader reader;
private static BufferedWriter out;
private static StringTokenizer tokenizer;
//private final static String filename = "filename";
/**
* call this method to initialize reader for InputStream and OututStream
*/
private static void init(InputStream input, OutputStream output) {
reader = new BufferedReader(new InputStreamReader(input));
out = new BufferedWriter(new OutputStreamWriter(output));
//reader = new BufferedReader(new FileReader(filename + ".in"));
//out = new BufferedWriter(new FileWriter(filename + ".out"));
tokenizer = new StringTokenizer("");
}
/**
* get next word
*/
private static String next() throws IOException {
while (!tokenizer.hasMoreTokens()) {
//TODO add check for eof if necessary
tokenizer = new StringTokenizer(reader.readLine());
}
return tokenizer.nextToken();
}
private static int nextInt() throws IOException {
return Integer.parseInt(next());
}
private static long nextLong() throws IOException {
return Long.parseLong(next());
}
private static double nextDouble() throws IOException {
return Double.parseDouble(next());
}
/**
* @param args the command line arguments
*/
public static void main(String[] args) throws IOException {
init(System.in, System.out);
int n = nextInt();
//long startTime = System.currentTimeMillis();
int k = nextInt();
int maxTot = (n * (n - 1)) / 2;
if (maxTot <= k) {
out.write("no solution\n");
} else {
for (int i = 0; i < n; i++) {
out.write("0 " + i + "\n");
}
}
//long runTime = System.currentTimeMillis() - startTime;
//out.write(runTime + "\n");
out.flush();
}
} | JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.util.Scanner;
public class Main2 {
public static void main(String args[]){
Scanner input = new Scanner(System.in);
int n = input.nextInt();
int k = input.nextInt();
sentence(n,k);
}
public static void sentence(int n,int k){
if(n == 2 || k >= (n*(n-1))/2){
System.out.println("no solution");
}else{
for(int i = 0 ; i < n ; i++){
System.out.println(0+" "+i);
}
}
}
} | JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | a,b=map(int,raw_input().split())
if b >= a*(a-1)/2:
print 'no solution'
else:
cnt = 0
for i in range(0,a):
print 0,i | PYTHON |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | import java.util.*;
import java.util.concurrent.LinkedBlockingDeque;
import java.util.zip.ZipEntry;
import javax.imageio.stream.MemoryCacheImageInputStream;
import javax.security.auth.Subject;
import javax.security.auth.kerberos.KerberosKey;
import javax.swing.plaf.basic.BasicScrollPaneUI.HSBChangeListener;
import javax.tools.JavaCompiler;
import javax.xml.bind.SchemaOutputResolver;
import javax.xml.crypto.dsig.spec.DigestMethodParameterSpec;
import org.omg.CORBA.TRANSACTION_MODE;
import org.omg.IOP.TAG_CODE_SETS;
import java.awt.font.GraphicAttribute;
import java.io.*;
import java.nio.channels.AcceptPendingException;
import java.rmi.server.RMIClassLoader;
import java.security.spec.DSAGenParameterSpec;
import java.sql.PseudoColumnUsage;
import java.time.Period;
//all inclusive
public class Main {
public static void main(String[] args) throws Exception {
int n=sc.nextInt();
int k=sc.nextInt();
if((n*n-n)/2<=k) {
pw.println("no solution");
}else {
for(int i=0;i<n;i++) {
pw.println(0+" "+i);
}
}
pw.close();
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(FileReader r) {
br = new BufferedReader(r);
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public double nextDouble() throws IOException {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if (x.charAt(0) == '-') {
neg = true;
start++;
}
for (int i = start; i < x.length(); i++)
if (x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
} else {
sb.append(x.charAt(i));
if (dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public long[] nextlongArray(int n) throws IOException {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public Long[] nextLongArray(int n) throws IOException {
Long[] a = new Long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public int[] nextIntArray(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public Integer[] nextIntegerArray(int n) throws IOException {
Integer[] a = new Integer[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public boolean ready() throws IOException {
return br.ready();
}
}
static class pair implements Comparable<pair> {
long x;
long y;
public pair(long x, long y) {
this.x = x;
this.y = y;
}
public String toString() {
return x + " " + y;
}
public boolean equals(Object o) {
if (o instanceof pair) {
pair p = (pair)o;
return p.x == x && p.y == y;
}
return false;
}
public int hashCode() {
return new Double(x).hashCode() * 31 + new Double(y).hashCode();
}
public int compareTo(pair other) {
if(this.x==other.x) {
return Long.compare(other.y, this.y);
}
return Long.compare(this.x, other.x);
}
}
static class tuble implements Comparable<tuble> {
int x;
int y;
int z;
public tuble(int x, int y, int z) {
this.x = x;
this.y = y;
this.z = z;
}
public String toString() {
return x + " " + y + " " + z;
}
public int compareTo(tuble other) {
if (this.x == other.x) {
if(this.y==other.y) {
return this.z - other.z;
}
return this.y - other.y;
} else {
return this.x - other.x;
}
}
}
public static long GCD(long a, long b) {
if (b == 0)
return a;
if (a == 0)
return b;
return (a > b) ? GCD(a % b, b) : GCD(a, b % a);
}
public static long LCM(long a, long b) {
return a * b / GCD(a, b);
}
static long Pow(long a, long e, long mod) // O(log e)
{
a %= mod;
long res = 1l;
while (e > 0) {
if ((e & 1) == 1)
res = (res * a) % mod;
a = (a * a) % mod;
e >>= 1l;
}
return res;
}
public static long modinverse(long a,long mod) {
return Pow(a, mod-2, mod);
}
static long nc(int n, int r) {
if (n < r)
return 0;
long v = fac[n];
v *= Pow(fac[r], mod - 2, mod);
v %= mod;
v *= Pow(fac[n - r], mod - 2, mod);
v %= mod;
return v;
}
public static boolean isprime(long a) {
if (a == 0 || a == 1) {
return false;
}
if (a == 2) {
return true;
}
for (int i = 2; i < Math.sqrt(a) + 1; i++) {
if (a % i == 0) {
return false;
}
}
return true;
}
public static boolean isPal(String s) {
boolean t = true;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) != s.charAt(s.length() - 1 - i)) {
t = false;
break;
}
}
return t;
}
public static long RandomPick(long[] a) {
int n = a.length;
int r = rn.nextInt(n);
return a[r];
}
public static int RandomPick(int[] a) {
int n = a.length;
int r = rn.nextInt(n);
return a[r];
}
public static void PH(String s, boolean reverse) {
prelen = s.length();
HashsArray[HashsArrayInd] = new int[prelen + 1];
prepow = new int[prelen];
if (HashsArrayInd == 0) {
int[] mods = { 1173017693, 1173038827, 1173069731, 1173086977, 1173089783, 1173092147, 1173107093,
1173114391, 1173132347, 1173144367, 1173150103, 1173152611, 1173163993, 1173174127, 1173204679,
1173237343, 1173252107, 1173253331, 1173255653, 1173260183, 1173262943, 1173265439, 1173279091,
1173285331, 1173286771, 1173288593, 1173298123, 1173302129, 1173308827, 1173310451, 1173312383,
1173313571, 1173324371, 1173361529, 1173385729, 1173387217, 1173387361, 1173420799, 1173421499,
1173423077, 1173428083, 1173442159, 1173445549, 1173451681, 1173453299, 1173454729, 1173458401,
1173459491, 1173464177, 1173468943, 1173470041, 1173477947, 1173500677, 1173507869, 1173522919,
1173537359, 1173605003, 1173610253, 1173632671, 1173653623, 1173665447, 1173675577, 1173675787,
1173684683, 1173691109, 1173696907, 1173705257, 1173705523, 1173725389, 1173727601, 1173741953,
1173747577, 1173751499, 1173759449, 1173760943, 1173761429, 1173762509, 1173769939, 1173771233,
1173778937, 1173784637, 1173793289, 1173799607, 1173802823, 1173808003, 1173810919, 1173818311,
1173819293, 1173828167, 1173846677, 1173848941, 1173853249, 1173858341, 1173891613, 1173894053,
1173908039, 1173909203, 1173961541, 1173968989, 1173999193};
mod = RandomPick(mods);
int[] primes = { 59, 61, 67, 71, 73, 79, 83, 89, 97, 101 };
prime = RandomPick(primes);
}
prepow[0] = 1;
if (!reverse) {
for (int i = 1; i < prelen; i++) {
prepow[i] = (int) ((1l * prepow[i - 1] * prime) % mod);
}
for (int i = 0; i < prelen; i++) {
if (s.charAt(i) >= 'a' && s.charAt(i) <= 'z')
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'a' + 1) * prepow[i]) % mod) % mod);
else
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'A' + 27) * prepow[i]) % mod) % mod);
}
} else {
for (int i = 1; i < prelen; i++) {
prepow[i] = (int) ((1l * prepow[i - 1] * prime) % mod);
}
for (int i = 0; i < prelen; i++) {
if (s.charAt(i) >= 'a' && s.charAt(i) <= 'z')
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'a' + 1) * prepow[prelen - 1 - i]) % mod) % mod);
else
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'A' + 27) * prepow[prelen - 1 - i]) % mod) % mod);
}
}
HashsArrayInd++;
}
public static int PHV(int l, int r, int n, boolean reverse) {
if (l > r) {
return 0;
}
int val = (int) ((1l * HashsArray[n - 1][r] + mod - HashsArray[n - 1][l - 1]) % mod);
if (!reverse) {
val = (int) ((1l * val * prepow[prelen - l]) % mod);
} else {
val = (int) ((1l * val * prepow[r - 1]) % mod);
}
return val;
}
public static void genprime(int n) {
boolean prime[] = new boolean[n+1];
for(int i=0;i<n;i++)
prime[i] = true;
for(int p = 2; p*p <=n; p++)
{
if(prime[p] == true)
{
for(int i = p*p; i <= n; i += p)
prime[i] = false;
}
}
for(int i = 2; i <= n; i++)
{
if(prime[i] == true) {
primes.put(i,primes.size());
primes2.put(primes2.size(),i);
}
}
}
public static long LSB(long x) {
return x&-x;
}
static class fenwick {
long[] arr;
public fenwick(Integer[] a) {
arr=new long[a.length+1];
for(int i=1;i<=a.length;i++) {
arr[i]+=a[i-1];
if(i+LSB(i)<=a.length) {
arr[(int) (i+LSB(i))]+=arr[i];
}
}
}
public fenwick(int[] a) {
arr=new long[a.length+1];
for(int i=1;i<=a.length;i++) {
arr[i]+=a[i-1];
if(i+LSB(i)<=a.length) {
arr[(int) (i+LSB(i))]+=arr[i];
}
}
}public fenwick(long[] a) {
arr=new long[a.length+1];
for(int i=1;i<=a.length;i++) {
arr[i]+=a[i];
if(i+LSB(i)<=a.length) {
arr[(int) (i+LSB(i))]+=arr[i];
}
}
}
public void update(int ind,long x) {
int i=ind;
while(i<arr.length) {
arr[i]+=x;
i+=LSB(i);
}
}
public long PrefixSum(int ind) {
long sum=0;
int i=ind;
while(i>0) {
sum+=arr[i];
i=(int) (i-LSB(i));
}
return sum;
}
public long RangeQuerey(int l,int r) {
return this.PrefixSum(r+1)-this.PrefixSum(l);
}
public long maxConsecutiveValue(int k) {
long max=Long.MIN_VALUE;
for(int i=k-1;i<arr.length-1;i++) {
max= Math.max(max, this.RangeQuerey(i-k+1, i));
}
return max;
}
public long minConsecutiveValue(int k) {
long min=Long.MAX_VALUE;
for(int i=k-1;i<arr.length-1;i++) {
min= Math.min(min, this.RangeQuerey(i-k+1, i));
}
return min;
}
public long value(int ind) {
return arr[ind];
}
}
static void sieveLinear(int N)
{
ArrayList<Integer> primes = new ArrayList<Integer>();
lp = new int[N + 1]; //lp[i] = least prime divisor of i
for(int i = 2; i <= N; ++i)
{
if(lp[i] == 0)
{
primes.add(i);
lp[i] = i;
}
int curLP = lp[i];
for(int p: primes)//all primes smaller than or equal my lowest prime divisor
if(p > curLP || p * 1l * i > N)
break;
else
lp[p * i] = p;
}
}
public static void primefactorization(int n) {
int x=n;
while(x>1) {
int lowestDivisor=lp[x];
while(x%lowestDivisor==0) {
primefactors.add(lowestDivisor);
x/=lowestDivisor;
}
}
}
public static class SuffixArray {
int[] SA;
int[] AS;
String SS;
public SuffixArray(String S) //has a terminating character (e.g. '$')
{
SS=S;
char[] s=new char[S.length()+1];
for(int i=0;i<S.length();i++) {
s[i]=S.charAt(i);
}
s[S.length()]='$';
int n = s.length, RA[] = new int[n];
SA = new int[n];
for(int i = 0; i < n; ++i) { RA[i] = s[i]; SA[i] = i; }
for(int k = 1; k < n; k <<= 1)
{
sort(SA, RA, n, k);
sort(SA, RA, n, 0);
int[] tmp = new int[n];
for(int i = 1, r = 0, s1 = SA[0], s2; i < n; ++i)
{
s2 = SA[i];
tmp[s2] = RA[s1] == RA[s2] && RA[s1 + k] == RA[s2 + k] ? r : ++r;
s1 = s2;
}
for(int i = 0; i < n; ++i)
RA[i] = tmp[i];
if(RA[SA[n-1]] == n - 1)
break;
}
AS=new int[SA.length];
for(int i=0;i<SA.length;i++) {
AS[SA[i]]=i;
}
}
public String toString() {
return Arrays.toString(SA);
}
public int get(int n) {
return SA[n];
}
public int Substring(String s) { // log(n)*|s|
int low=0;
int high=SA.length;
int mid=(low+high)/2;
int ind=-1;
while(low<high-1) {
if(SS.length()-SA[mid]<s.length()) {
boolean less=false;
for(int i=SA[mid];i<SS.length();i++) {
if(SS.charAt(i)>s.charAt(i-SA[mid])) {
less=true;
break;
}
if(SS.charAt(i)<s.charAt(i-SA[mid])) {
less=false;
break;
}
}
if(!less) {
low=mid;
}else {
high=mid;
}
}else {
boolean less=true;
boolean equal=true;
for(int i=SA[mid];i<SA[mid]+s.length();i++) {
if(SS.charAt(i)<s.charAt(i-SA[mid])&&equal) {
less=false;
equal=false;
break;
}
if(SS.charAt(i)!=s.charAt(i-SA[mid])){
equal=false;
}
}
if(equal) {
ind=SA[mid];
}
if(!less) {
low=mid;
}else {
high=mid;
}
}
mid=(low+high)/2;
}
return ind;
}
public int LastSubstring(String s) { // log(n)*|s|
int low=0;
int high=SA.length;
int mid=(low+high)/2;
int ind=-1;
while(low<high-1) {
if(SS.length()-SA[mid]<s.length()) {
boolean less=true;
for(int i=SA[mid];i<SS.length();i++) {
if(SS.charAt(i)<s.charAt(i-SA[mid])) {
break;
}
if(SS.charAt(i)>s.charAt(i-SA[mid])) {
less=false;
break;
}
}
if(less) {
low=mid;
}else {
high=mid;
}
}else {
boolean less=true;
boolean equal=true;
for(int i=SA[mid];i<SA[mid]+s.length();i++) {
if(SS.charAt(i)>s.charAt(i-SA[mid])&&equal) {
less=false;
equal=false;
break;
}
if(SS.charAt(i)!=s.charAt(i-SA[mid])){
equal=false;
}
}
if(equal) {
ind=SA[mid];
}
if(less) {
low=mid;
}else {
high=mid;
}
}
mid=(low+high)/2;
}
return ind;
}
public int CountSubstring(String s) {
int z=LastSubstring(s);
if(z==-1)
return 0;
return AS[z]-AS[Substring(s)]+1;
}
public void sort(int[] SA, int[] RA, int n, int k)
{
int maxi = Math.max(256, n), c[] = new int[maxi];
for(int i = 0; i < n; ++i)
c[i + k < n ? RA[i + k] : 0]++;
for(int i = 0, sum = 0; i < maxi; ++i)
{
int t = c[i];
c[i] = sum;
sum += t;
}
int[] tmp = new int[n];
for(int i = 0; i < n; ++i)
{
int j = SA[i] + k;
tmp[c[j < n ? RA[j] : 0]++] = SA[i];
}
for(int i = 0; i < n; ++i)
SA[i] = tmp[i];
}
}
static LinkedList<Integer>primefactors=new LinkedList<>();
static TreeMap<Integer,Integer>primes=new TreeMap<Integer, Integer>();
static TreeMap<Integer,Integer>primes2=new TreeMap<Integer, Integer>();
static int[]lp;
static int[][] HashsArray;
static int HashsArrayInd = 0;
static int[] prepow;
static int prelen = 0;
static int prime = 61;
static long fac[];
static int mod = 1000000007;
static Random rn = new Random();
static Scanner sc = new Scanner(System.in);
static PrintWriter pw = new PrintWriter(System.out);
} | JAVA |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 100500;
const int infi = (1 << 30);
const long long infl = (1LL << 62);
const double eps = 1e-9;
const long long mod = 1000000007LL;
const double pi = acos(-1.0);
int n, k;
int main() {
cin >> n >> k;
if (n * (n - 1) <= k * 2) {
cout << "no solution\n";
} else {
for (int i = 0; i < n; i++) {
cout << 0 << " " << i << endl;
}
}
return 0;
}
| CPP |
312_C. The Closest Pair | Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
scanf("%d %d", &n, &k);
if ((n * (n - 1) / 2 <= k))
printf("no solution\n");
else {
for (int i = 0; i < n; i++) printf("0 %d\n", i);
}
return 0;
}
| CPP |
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