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| description
stringlengths 29
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| source
int64 1
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| difficulty
int64 0
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| solution
stringlengths 7
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| language
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287_C. Lucky Permutation | A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int n;
int a[100010];
void output() {
for (int i = 0; i < n; i++) printf("%d ", a[i]);
puts("");
}
int check() {
for (int i = 0; i < n; i++)
if (a[a[i] - 1] != n - i) return 0;
return 1;
}
int main() {
cin >> n;
for (int i = 0; i < n / 2; i += 2) a[i] = i + 2;
for (int i = 1; i < n / 2; i += 2) a[i] = n + 1 - i;
int j = 1;
for (int i = n - 2; i >= n / 2; i -= 2, j += 2) a[i] = j;
j = n - 1;
for (int i = n - 1; i >= n / 2; i -= 2, j -= 2) a[i] = j;
if (n & 1) a[n / 2] = n / 2 + 1;
if (check() == 1)
output();
else
printf("-1\n");
return 0;
}
| CPP |
287_C. Lucky Permutation | A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4 | 2 | 9 | import java.util.*;
import java.math.*;
import java.io.*;
public class Main
{
public static void main(String args[]) throws IOException
{
InputReader in=new InputReader(System.in);
PrintWriter out=new PrintWriter(System.out);
int N=in.readInt();
int A[]=new int[N+1];
Arrays.fill(A,-1);
if(N%4!=0&&N%4!=1)
{
System.out.println("-1");
return;
}
if(N==1)
{
System.out.println("1");
return;
}
TreeSet<Integer> T=new TreeSet<Integer>();
for(int i=1;i<=N;i++)
T.add(i);
if(N%4==1)
{
A[(N+1)/2]=(N+1)/2;
T.remove((N+1)/2);
}
for(int i=1;i<=N;i++)
{
if(A[i]==-1)
{
int ai=T.pollFirst();
if(ai==i)
{
int temp=T.pollFirst();
T.add(ai);
ai=temp;
}
A[i]=ai;
//System.out.println(i+" gets "+ai);
A[ai]=N-i+1;
//System.out.println(ai+" gets "+(N-i+1));
A[N-i+1]=N-ai+1;
//System.out.println((N-i+1)+" gets "+(N-ai+1));
A[N-ai+1]=i;
//System.out.println((N-ai+1)+" gets "+i);
T.remove(ai);
T.remove(N-i+1);
T.remove(N-ai+1);
T.remove(i);
}
}
check(A);
for(int i=1;i<=N;i++)
out.print(A[i]+" ");
out.println();
out.close();
}
private static void check(int[] a)
{
int N=a.length-1;
for(int i=1;i<=N;i++)
{
if(a[a[i]]!=N-i+1)
{
System.out.println((1/0));
}
}
}
}
class permGen
{
public int A[];
public permGen(int a[])
{
A=new int[a.length];
for(int i=0;i<a.length;i++)
A[i]=a[i];
}
public int[] nextPerm()
{
//Find the largest index k such that a[k] < a[k + 1]. If no such index exists, the permutation is the last permutation.
int N=A.length;
int k=-1;
for(int i=0;i<N-1;i++)
{
if(A[i]<A[i+1])
k=i;
}
if(k==-1)
return null;
//Find the largest index l such that a[k] < a[l]. Since k + 1 is such an index, l is well defined and satisfies k < l
int l=-1;
for(int i=k+1;i<N;i++)
{
if(A[k]<A[i])
l=i;
}
//Swap a[k] with a[l].
swap(A,k,l);
//Reverse the sequence from a[k + 1] up to and including the final element a[n]
int i=k+1;
int j=N-1;
while (i<=j)
{
swap(A,i,j);
i++;
j--;
}
return A;
}
private void swap(int A[],int i,int j)
{
int t=A[i];
A[i]=A[j];
A[j]=t;
}
}
class InputReader
{
private InputStream stream;
private byte[] buf=new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream)
{
this.stream=stream;
}
public 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++];
}
public int readInt()
{
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 boolean isSpaceChar(int c)
{
if(filter!=null)
return filter.isSpaceChar(c);
return isWhitespace(c);
}
public static boolean isWhitespace(int c)
{
return c==' '||c=='\n'||c=='\r'||c=='\t'||c==-1;
}
public char readCharacter()
{
int c=read();
while (isSpaceChar(c))
c=read();
return (char) c;
}
public interface SpaceCharFilter
{
public boolean isSpaceChar(int ch);
}
}
| JAVA |
287_C. Lucky Permutation | A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long p[100001];
int main(void) {
ios::sync_with_stdio(false);
cin.tie(NULL);
long n;
cin >> n;
if (n % 4 == 2 || n % 4 == 3)
cout << "-1";
else {
if (n % 2) p[n / 2 + 1] = n / 2 + 1;
for (int i = 1; i < n / 2; i += 2) {
p[i] = i + 1;
p[i + 1] = n - i + 1;
p[n - i] = i;
p[n - i + 1] = n - i;
}
for (int i = 1; i <= n; i++) cout << p[i] << " ";
}
return 0;
}
| CPP |
287_C. Lucky Permutation | A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4 | 2 | 9 | import java.util.*;
import java.io.*;
public class Round176A {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
PrintWriter pw = new PrintWriter(System.out);
int n = sc.nextInt();
if(n%4==2 || n%4==3){
System.out.println(-1);
return;
}
if(n%4==0){
int[] p = new int[n+1];
int k = n/4;
for(int i=1; i<=k; i++) p[2*i-1] = 2*i;
for(int i=1; i<=k; i++) p[2*i] = 4*k+2-2*i;
for(int i=2*k+1; i<=4*k; i++) p[i] = 4*k+1 - p[4*k+1-i];
for(int i=1; i<=n; i++) pw.print(p[i]+" ");
pw.println();
pw.flush();
return;
}
int[] p = new int[n+1];
int k = n/4;
for(int i=1; i<=k; i++) p[2*i-1] = 2*i;
for(int i=1; i<=k; i++) p[2*i] = 4*k+3-2*i;
for(int i=2*k+2; i<=4*k+1; i++) p[i] = 4*k+2-p[4*k+2-i];
p[2*k+1] = 2*k+1;
for(int i=1; i<=n; i++) pw.print(p[i]+" ");
pw.println();
pw.flush();
}
}
| JAVA |
287_C. Lucky Permutation | A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int n;
cin >> n;
if (n % 4 > 1) {
cout << "-1\n";
return 0;
}
int p[n];
if (n % 4 == 1) {
p[n / 2] = (n + 1) / 2;
}
if (n != 1) {
p[0] = 2;
p[1] = n;
p[n - 2] = 1;
p[n - 1] = n - 1;
}
for (int i = 2, j = n - 4; i < n / 2; i += 2, j -= 2) {
p[i] = p[i - 2] + 2;
p[i + 1] = p[i - 1] - 2;
p[j] = p[i] - 1;
p[j + 1] = p[i + 1] - 1;
}
for (int i : p) cout << i << " ";
cout << endl;
return 0;
}
| CPP |
287_C. Lucky Permutation | A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4 | 2 | 9 | n=int(input())
if n==1:
print (1)
exit()
if n%4>1:
print (-1)
exit()
ans=[-1]*n
left=n
start=n-2
nums=1
nume=n
while left>=4:
ans[start]=nums
ans[nums-1]=nums+1
ans[nums]=nume
ans[nume-1]=nume-1
start-=2
nums+=2
nume-=2
left-=4
# print (ans)
if left==1:
ans[start+1]=start+2
print (*ans) | PYTHON3 |
287_C. Lucky Permutation | A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int a[100001];
int main() {
int n;
cin >> n;
if (n % 4 == 2 || n % 4 == 3) {
cout << -1 << endl;
return 0;
}
int i;
if (n % 2 == 1) a[(n + 1) / 2] = (n + 1) / 2;
for (i = 1; i <= n / 2;) {
a[i] = i + 1;
a[i + 1] = n - i + 1;
a[n - i + 1] = n - i;
a[n - i] = i;
i += 2;
}
for (i = 1; i <= n; i++) {
cout << a[i] << " ";
}
cout << endl;
return 0;
}
| CPP |
287_C. Lucky Permutation | A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4 | 2 | 9 | import java.util.*;
import java.lang.*;
import java.math.*;
import java.io.*;
import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;
// Lucky Permutation
// 2013/03/23
public class P286A{
Scanner sc=new Scanner(System.in);
int n;
void run(){
n=sc.nextInt();
solve();
}
void solve(){
if(n%4==0||n%4==1){}else{
println("-1");
return;
}
TreeSet<Integer> set=new TreeSet<Integer>();
for(int i=0; i<n; i++){
set.add(i);
}
TreeSet<Integer> temp=new TreeSet<Integer>();
int[] a=new int[n];
fill(a, -1);
for(int index=0; index<n/2; index++){
if(a[index]>-1){
continue;
}
int i;
for(i=set.pollFirst(); index==i||n-1-index==i; i=set.pollFirst()){
temp.add(i);
}
a[index]=i;
a[n-1-index]=n-1-i;
a[a[index]]=n-1-index;
a[a[n-1-index]]=index;
set.addAll(temp);
set.remove(i);
set.remove(n-1-i);
set.remove(n-1-index);
set.remove(index);
temp.clear();
}
if(n%2==1){
a[n/2]=n/2;
}
StringBuilder sb=new StringBuilder();
for(int i=0; i<n; i++){
sb.append(a[i]+1);
if(i<n-1){
sb.append(' ');
}
}
println(sb.toString());
}
void println(String s){
System.out.println(s);
}
public static void main(String[] args){
Locale.setDefault(Locale.US);
new P286A().run();
}
}
| JAVA |
287_C. Lucky Permutation | A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4 | 2 | 9 | import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.util.StringTokenizer;
/**
* Built using CHelper plug-in
* Actual solution is at the top
* @author PM
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskA solver = new TaskA();
solver.solve(1, in, out);
out.close();
}
}
class TaskA {
public void solve(int testNumber, InputReader in, PrintWriter out) {
int n = in.nextInt();
if (n%4==2 || n%4==3) {out.println(-1); return;}
int[] res = new int[n];
for (int i = 0; i < n; i++) {
res[i] = i+1;
}
for (int at = 0; at < n/2; at +=2) {
int last = (n-1-at);
int first = res[at];
res[at] = res[at+1];
res[at+1] = res[last];
res[last] = res[last-1];
res[last-1] = first;
}
for (int i = 0; i < n; i++) {
out.print(res[i] + " ");
}
}
}
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 nextInt() {
return Integer.parseInt(next());
}
}
| JAVA |
287_C. Lucky Permutation | A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4 | 2 | 9 | import java.util.Scanner;
public class CodeforcesRound176C {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner kde= new Scanner(System.in);
int n=kde.nextInt();
if((n%4==2)||(n%4==3))
{
System.out.println(-1);
return;
}
int[] m= new int[n+1];
if (n % 4 == 0)
{
int t = n/4;
for (int i = 1; i <= t; i++)
{
int odd = 2*i - 1;
int even = 2*i;
m[odd] = even;
m[even] = n+1 - odd;
m[n+1 - odd] = n+1 - even;
m[n+1 - even] = odd;
}
}
if (n % 4 == 1)
{
int t = (n - 1)/4;
for (int i = 1; i <= t; i++)
{
int odd = 2*i-1;
int even = 2*i;
m[odd] = even;
m[even] = n+1 - odd;
m[n+1 - odd] = n+1 - even;
m[n+1 - even] = odd;
}
m[(n+1)/2] = (n+1)/2;
}
for (int i = 1; i <= n; i++)
{
System.out.print(m[i]+" ");
}
System.out.println();
}
}
| JAVA |
287_C. Lucky Permutation | A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
inline bool xdy(double x, double y) { return x > y + 1e-9; }
inline bool xddy(double x, double y) { return x > y - 1e-9; }
inline bool xcy(double x, double y) { return x < y - 1e-9; }
inline bool xcdy(double x, double y) { return x < y + 1e-9; }
const long long int mod = 1000000007;
int n;
int ans[100005];
int main() {
ios_base::sync_with_stdio(0);
while (~scanf("%d", &n)) {
if (n % 4 == 2 || n % 4 == 3) {
printf("-1\n");
continue;
}
for (int i = 1; i <= n / 2; i += 2) {
ans[i] = i + 1;
ans[i + 1] = n - i + 1;
ans[n - i + 1] = n - i;
ans[n - i] = i;
}
if (n & 1) ans[n / 2 + 1] = n / 2 + 1;
for (int i = 1; i <= n; i++) {
printf("%d ", ans[i]);
}
printf("\n");
}
}
| CPP |
287_C. Lucky Permutation | A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
if (n % 4 > 1) {
cout << -1;
return 0;
}
int m = n / 4, perm[100001];
for (int i = 0; i < m; i++) {
int a, b, c, d;
a = 2 * i + 1;
b = 2 * i + 2;
c = n - 2 * i;
d = n - 2 * i - 1;
perm[a] = b;
perm[b] = c;
perm[c] = d;
perm[d] = a;
}
if (n % 4 == 1) perm[n / 2 + 1] = n / 2 + 1;
for (int i = 1; i <= n; i++) {
cout << perm[i] << " ";
}
}
| CPP |
287_C. Lucky Permutation | A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4 | 2 | 9 |
import java.util.*;
import java.io.*;
public class Main {
public static void main(String[] args) throws Exception {
// Scanner input = new Scanner(new BufferedReader(new InputStreamReader(System.in), 16000));
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream), 64000));
// 2 8 4 6 3 5 1 7
//
// 13 12 11 10 9 8 7 6 5 4 3 2 1
// 1 2 3 4 5 6 7 8 9 10 11 12 13
// 2 13 4 11 6 9 7 5 8 3 10 1 12
int n = in.nextInt();
boolean odd = false ;
int oddV = 0 ;
if(n % 2 != 0) {
odd = true ;
oddV = 1 ;
n-- ;
}
if(n % 4 != 0){
out.println(-1);
out.close();
return ;
}
for(int i = 2 ; i <= n / 2 ; i+=2){
out.println(i );
out.println(n + 2 + oddV - i );
}
if(odd) out.println((n/ 2) + 1);
for(int i = n / 2 + 1 ; i <= n - 1 ; i += 2) {
out.println(n - i);
out.println(i + oddV);
}
out.close();
}
}
class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 64000);
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 long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
| JAVA |
287_C. Lucky Permutation | A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4 | 2 | 9 |
import java.io.PrintWriter;
import java.util.Scanner;
public class C {
public static void main(String [] args){
Scanner in = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
int n = in.nextInt();
if(n%4==1 || n%4==0){
int a [] = new int[n+1];
for(int i = 1 ; i <= n/2 ; i+=2 ){
a[i]=i+1;
a[n-i+1] = n-i;
}
for(int i = 2 ; i <= n/2 ; i+=2){
a[i]=n-i+2;
a[n-i+1] = i - 1;
}
if(n%2 ==1){
a[n/2+1]= n/2+1;
}
out.print(a[1]);
for(int i = 2 ; i <= n ; i++){
out.print(" "+a[i]);
}
}
else{
out.print(-1);
}
out.close();
}
}
| JAVA |
287_C. Lucky Permutation | A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4 | 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.BufferedReader;
import java.util.LinkedList;
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;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskA solver = new TaskA();
solver.solve(1, in, out);
out.close();
}
static class TaskA {
public void solve(int testNumber, InputReader in, PrintWriter out) {
int n = in.nextInt();
if (n % 4 < 2) {
LinkedList<Integer> ans = new LinkedList<>();
int curN = 0;
if (n % 4 == 1) {
ans.add(1);
curN = 1;
} else {
ans.add(2);
ans.add(4);
ans.add(1);
ans.add(3);
curN = 4;
}
while (ans.size() < n) {
ans.addFirst(curN + 4);
ans.addFirst(2);
ans.addLast(1);
ans.addLast(curN + 3);
curN += 4;
}
int[] ansArr = new int[n];
int k = 0;
for (int i : ans) {
ansArr[k] = i;
k++;
}
for (int i = 0; i < n / 4; i++) {
ansArr[2 * i] += 2 * i;
ansArr[2 * i + 1] += 2 * i;
ansArr[n - 1 - 2 * i] += 2 * i;
ansArr[n - 2 - 2 * i] += 2 * i;
}
if (n % 4 == 1) {
ansArr[n / 2] += 2 * (n / 4);
}
for (int i : ansArr) {
out.print(i + " ");
}
} else {
out.println(-1);
}
}
}
static class InputReader {
private StringTokenizer tokenizer;
private BufferedReader reader;
public InputReader(InputStream inputStream) {
reader = new BufferedReader(new InputStreamReader(inputStream));
}
private void fillTokenizer() {
if (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (Exception e) {
throw new RuntimeException(e);
}
}
}
public String next() {
fillTokenizer();
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}
| JAVA |
287_C. Lucky Permutation | A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int n;
deque<int> ans;
bool used[100500];
int add(int i, int n) {
if (n % 2 == 0) {
if (i >= n / 2) i = n - i - 1;
return i / 2;
}
if (i > n / 2) i = n - i - 1;
return i / 2;
}
int main() {
cin >> n;
if (n == 1) {
cout << 1;
return 0;
} else if (n % 4 == 2 || n % 4 == 3) {
cout << -1;
return 0;
}
if (n % 4 == 1) {
ans.push_back(1);
while (ans.size() < n) {
int sz = ans.size();
ans.push_front(sz + 4);
ans.push_front(2);
ans.push_back(1);
ans.push_back(sz + 3);
}
for (int i = 0; i < ans.size() / 2; i++) ans[i] += 2 * (i / 2);
for (int i = ans.size() / 2; i < ans.size(); i++)
ans[i] += 2 * ((ans.size() - i - 1) / 2);
} else {
ans.push_back(2);
ans.push_back(4);
ans.push_back(1);
ans.push_back(3);
while (ans.size() < n) {
int sz = ans.size();
ans.push_front(sz + 4);
ans.push_front(2);
ans.push_back(1);
ans.push_back(sz + 3);
}
for (int i = 0; i < ans.size() / 2; i++) ans[i] += 2 * (i / 2);
for (int i = ans.size() / 2; i < ans.size(); i++)
ans[i] += 2 * ((ans.size() - i - 1) / 2);
}
for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
return 0;
}
| CPP |
287_C. Lucky Permutation | A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4 | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int a[maxn];
int main() {
int n;
cin >> n;
if (n % 4 == 2 || n % 4 == 3)
cout << -1 << endl;
else if (n % 4 == 0) {
for (int i = 1; i <= (n / 2); i++) {
if ((i % 2) == 1)
a[i] = i + 1;
else
a[i] = n - i + 2;
}
for (int i = n; i > (n / 2); i--) {
if ((i % 2) == (n % 2))
a[i] = i - 1;
else
a[i] = n - i;
}
for (int i = 1; i <= n; i++) cout << a[i] << " ";
cout << endl;
} else {
for (int i = 1; i <= (n / 2); i++) {
if ((i % 2) == 1)
a[i] = i + 1;
else
a[i] = n - i + 2;
}
for (int i = n; i > ((n + 1) / 2); i--) {
if ((i % 2) == (n % 2))
a[i] = i - 1;
else
a[i] = n - i;
}
a[(n + 1) / 2] = (n + 1) / 2;
for (int i = 1; i <= n; i++) cout << a[i] << " ";
cout << 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())
if n*(n-1) <= k*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 | # -*- coding: utf-8 -*-
"""
Created on Sat Jul 06 18:16:44 2013
@author: workshop
"""
from __future__ import division;
from bisect import *;
from fractions import Fraction;
import sys;
from math import *;
from fractions import *;
import io;
import re;
INF = 987654321987654321987654321;
def readint(delimiter=' ') :
return map(int, raw_input().split(delimiter));
def readstr(delimiter=' ') :
return raw_input().split(delimiter);
def readfloat(delimiter=' ') :
return map(float, raw_input().split(delimiter));
def index(a, x):
'Locate the leftmost value exactly equal to x'
i = bisect_left(a, x)
if i != len(a) and a[i] == x:
return i
raise ValueError
def find_lt(a, x):
'Find rightmost value less than x'
i = bisect_left(a, x)
if i:
return a[i-1]
raise ValueError
def find_le(a, x):
'Find rightmost value less than or equal to x'
i = bisect_right(a, x)
if i:
return a[i-1]
raise ValueError
def find_gt(a, x):
'Find leftmost value greater than x'
i = bisect_right(a, x)
if i != len(a):
return a[i]
raise ValueError
def find_ge(a, x):
'Find leftmost item greater than or equal to x'
i = bisect_left(a, x)
if i != len(a):
return a[i]
raise ValueError
def bin_search(a, x, left, right) :
while left<=right :
mid = (left + right)//2;
if a[mid] == x :
return mid;
elif a[mid] < x :
left = mid + 1;
elif a[mid] > x :
right = mid - 1;
pass
return -1;
pass
def printf(format, *args):
"""Format args with the first argument as format string, and write.
Return the last arg, or format itself if there are no args."""
sys.stdout.write(str(format) % args)
pass
if __name__ == '__main__':
n, k = readint();
if (n-1)*(n)/2 <= k :
print "no solution";
else :
for ii in xrange(n) :
print 0, ii;
pass
pass
pass | 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 | /*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
//package Practice;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
/**
*
* @author Rohan
*/
public class Main {
/**
* @param args the command line arguments
*/
public static void main(String[] args) throws IOException {
// TODO code application logic here
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n,k;
String[] S=br.readLine().split(" ");
n=Integer.parseInt(S[0]);
k=Integer.parseInt(S[1]);
if(((n*(n-1))/2)<=k)
System.out.println("no solution");
else{
int temp=1000000;
for(int i=0;i<n;i++){
System.out.println(0 + " " + temp);
temp+=(2*i+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 | '''
Created on 2013-5-27
@author: zhxfl
'''
a, b = map(int,raw_input().split());
n = a * (a - 1) / 2;
if b >= n:
print 'no solution'
else:
for i in range(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 | /*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
import java.io.*;
import java.math.BigInteger;
import java.util.*;
import java.text.*;
public class cf312C {
static BufferedReader br;
static Scanner sc;
static PrintWriter out;
public static void initA() {
try {
br = new BufferedReader(new InputStreamReader(System.in));
//br = new BufferedReader(new FileReader("input.txt"));
sc = new Scanner(System.in);
//out = new PrintWriter("output.txt");
out = new PrintWriter(System.out);
} catch (Exception e) {
}
}
static boolean next_permutation(Integer[] p) {
for (int a = p.length - 2; a >= 0; --a) {
if (p[a] < p[a + 1]) {
for (int b = p.length - 1;; --b) {
if (p[b] > p[a]) {
int t = p[a];
p[a] = p[b];
p[b] = t;
for (++a, b = p.length - 1; a < b; ++a, --b) {
t = p[a];
p[a] = p[b];
p[b] = t;
}
return true;
}
}
}
}
return false;
}
public static void initB() {
try {
br = new BufferedReader(new FileReader("input.txt"));
sc = new Scanner(new FileReader("input.txt"));
out = new PrintWriter("output.txt");
} catch (Exception e) {
}
}
public static String getString() {
try {
return br.readLine();
} catch (Exception e) {
}
return "";
}
public static Integer getInt() {
try {
return Integer.parseInt(br.readLine());
} catch (Exception e) {
}
return 0;
}
public static Integer[] getIntArr() {
try {
StringTokenizer temp = new StringTokenizer(br.readLine());
int n = temp.countTokens();
Integer temp2[] = new Integer[n];
for (int i = 0; i < n; i++) {
temp2[i] = Integer.parseInt(temp.nextToken());
}
return temp2;
} catch (Exception e) {
}
return null;
}
public static Long[] getLongArr() {
try {
StringTokenizer temp = new StringTokenizer(br.readLine());
int n = temp.countTokens();
Long temp2[] = new Long[n];
for (int i = 0; i < n; i++) {
temp2[i] = Long.parseLong(temp.nextToken());
}
return temp2;
} catch (Exception e) {
}
return null;
}
public static String[] getStringArr() {
try {
StringTokenizer temp = new StringTokenizer(br.readLine());
int n = temp.countTokens();
String temp2[] = new String[n];
for (int i = 0; i < n; i++) {
temp2[i] = (temp.nextToken());
}
return temp2;
} catch (Exception e) {
}
return null;
}
public static int getMax(Integer[] ar) {
int t = ar[0];
for (int i = 0; i < ar.length; i++) {
if (ar[i] > t) {
t = ar[i];
}
}
return t;
}
public static void print(Object a) {
out.println(a);
}
public static void print(String s, Object... a) {
out.printf(s, a);
}
public static int nextInt() {
return sc.nextInt();
}
public static double nextDouble() {
return sc.nextDouble();
}
public static void main(String[] ar) {
initA();
solve();
out.flush();
}
static long hitung(long l){
long byk_el = l-1;
if(byk_el%2==0){
return (byk_el+1) * (byk_el/2);
}else{
return ((byk_el+1)/2) * (byk_el);
}
}
public static void solve() {
Long xx[] = getLongArr();
long n = xx[0], tle = xx[1];
long milyar = 1000000000l;
long maksimal = hitung(n);
if(maksimal<=tle){
//print(maksimal);
print("no solution");
return;
}
long needed ;
for(int i =1 ;i <=n ;i++){
print(i+" "+i*5000);
}
}
}
| 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 GCD(int a, int b) {
if (!a) return b;
return GCD(b % a, a);
}
vector<int> x(4000);
vector<int> y(4000);
void D(double x, double y, double x1, double y1) {
double d = (x - x1) * (x - x1) + (y - y1) * (y - y1);
cout << sqrt(d) << "\n";
}
int main() {
int n, k;
cin >> n >> k;
int tot = ((n - 1) * n) / 2;
if (tot > k) {
y[0] = 0;
for (int i = 1; i < n; i++) {
y[i] = y[i - 1] + (100000 - i);
}
for (int i = 0; i < n; i++) {
printf("%d %d\n", i, y[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.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.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;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskC solver = new TaskC();
solver.solve(1, in, out);
out.close();
}
static class TaskC {
public void solve(int testNumber, InputReader in, PrintWriter out) {
int n = in.nextInt();
int k = in.nextInt();
if (k >= (n * (n - 1)) / 2) {
out.println("no solution");
return;
}
for (int i = 0; i < n; i++) {
out.println(0 + " " + i);
}
}
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream));
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());
}
}
}
| 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 i, n, k;
cin >> n >> k;
if (n * (n - 1) / 2 <= k) {
puts("no solution");
return 0;
}
for (i = 1; 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() {
long n, k;
cin >> n >> k;
long kc = 2001;
if (k >= n * (n - 1) / 2) {
cout << "no solution";
return 0;
}
long sum = 0;
long ax[2005], ay[2005];
for (int i = 1; i <= n; i++) {
sum += kc;
ax[i] = 0;
ay[i] = sum;
cout << 0 << " " << sum << endl;
--kc;
}
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;
template <class T>
T gmin(T u, T v) {
return (u < v) ? u : v;
}
template <class T>
T gmax(T u, T v) {
return (u > v) ? u : v;
}
template <class T>
T gcd(T u, T v) {
if (v == 0) return u;
return (u % v == 0) ? v : gcd(v, u % v);
}
int main() {
long long n, m, tmp;
cin >> n >> m;
tmp = n * (n - 1) / 2;
if (tmp <= m) {
puts("no solution");
return 0;
}
for (m = 1, tmp = 0; m <= n; ++m, tmp += n + 10)
cout << m << " " << tmp << 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;
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;
scanf("%d %d", &n, &k);
if ((n * (n - 1)) / 2 <= k) {
printf("no solution\n");
return 0;
}
for (int i = 0; i < (int)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.*;
public class Pair{
public static void main(String[] args){
Scanner reader = new Scanner(System.in);
int n = reader.nextInt();
int k = reader.nextInt();
Point[] p = new Point[n];
Point dir = new Point(1,2001);
p[0] = new Point(0,0);
for(int i = 1; i < n; i++)
p[i] = p[i-1].add(dir);
if((n*(n-1))/2 > k){
for(int i = 0; i < n; i++)
System.out.println(p[i].x + " " + p[i].y);
}else{
System.out.println("no solution");
}
}
public static boolean test(Point[] p, int k){
Arrays.sort(p);
int n = p.length;
int tot = 0;
double d = 1e16;
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, p[i].dis(p[j]));
}
return tot > k;
}
public static class Point implements Comparable<Point>{
int x,y;
public Point(int _x, int _y){
x = _x;
y = _y;
}
public Point add(Point p){
return new Point(x+p.x, y+p.y);
}
public double dis(Point p){
return Math.sqrt(Math.pow(x-p.x,2)+Math.pow(y-p.y,2));
}
public int compareTo(Point p){
if(x == p.x)
return y-p.y;
return x-p.x;
}
}
}
| 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 n, k;
int main() {
cin >> n >> k;
if (n * (n - 1) / 2 <= k) {
cout << "no solution\n";
return 0;
}
for (int i = 1; i <= n; i++) {
cout << 1 << " " << i + 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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
if (((n) * (n - 1)) / 2 > k) {
for (int i = 0; i < n; i++) cout << "0 " << i << endl;
} else {
cout << "no solution" << 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 - 1) * n / 2 <= k) {
cout << "no solution" << endl;
} else {
int tot = 0;
int i, j;
for (i = 0; i <= 1e9; i++)
for (j = 0; j <= 1e9; j++) {
if (tot < n) {
cout << i << " " << j << endl;
tot++;
} else
j = 1e9, i = 1e9;
}
}
}
| 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" << endl;
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=list(map(int,input().split()))
if (n*n-n)//2<=k:
print("no solution")
else:
x=0
y=0
store=[]
count=0
store.append(str(x)+' '+str(y))
while len(store)<n:
y+=1
store.append(str(x)+' '+str(y))
for j in store:
print(j) | 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 n, k;
int main() {
cin >> n >> k;
if (((n * (n - 1)) >> 1) <= k) {
cout << "no solution";
return 0;
}
for (int i = 0; i < n; ++i) {
cout << i << ' ' << 1000000000 - i * 3000 << 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;
while (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;
}
}
}
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 n, k;
cin >> n >> k;
if (k >= n * (n - 1) / 2) {
cout << "no solution\n";
} else {
for (long long 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 | #include <bits/stdc++.h>
using namespace std;
long long Min(long long i, long long j) { return i < j ? i : j; }
long long Max(long long i, long long j) { return i > j ? i : j; }
int main() {
long long a, b, c, d, e, i, j, k, l, m, n;
while (cin >> n >> k) {
if (k >= (n * (n - 1)) / 2)
cout << "no solution\n";
else {
a = -2000000;
for (i = 1; i <= n; i++) cout << 0 << ' ' << ++a << 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() {
ios_base::sync_with_stdio(false);
int n, k;
cin >> n >> k;
if (n * (n - 1) / 2 <= k) {
cout << "no solution" << '\n';
return 0;
} else {
for (int i = 1; i <= n; i++) {
cout << "1 " << i + 1 << 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.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class CF {
public static void main(String[] args) throws IOException {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
int n = in.nextInt();
int k = in.nextInt();
if (n * (n - 1) / 2 <= k) {
out.println("no solution");
} else {
int x = 0;
int y = 0;
for (int i = 0; i < n; i++) {
out.println(x + " " + y);
y += (n - x);
x++;
}
}
out.close();
}
}
class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream));
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());
}
} | 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 y, k, n, x, i;
scanf("%d%d", &n, &k);
n--;
x = (n * (n + 1)) / 2;
if (x > k) {
for (i = 1; i <= n + 1; i++) printf("0 %d\n", i);
} else
printf("no solution\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 | import java.util.Arrays;
import java.util.Locale;
import java.util.Scanner;
public class NearestPairSolver {
private int n;
private int k;
public static void main(String[] args) {
NearestPairSolver solver = new NearestPairSolver();
solver.readData();
int[] solution = solver.solve();
if (solution[0] == -1) {
solver.print("no solution");
} else {
solver.print(solution);
}
}
private void print(int[] values) {
StringBuilder builder = new StringBuilder();
for (int value : values) {
builder.append("0 ");
builder.append(value);
builder.append(System.getProperty("line.separator"));
}
print(builder);
}
private void print(Object value) {
System.out.println(value);
}
private void readData() {
Scanner scanner = new Scanner(System.in);
n = scanner.nextInt();
k = scanner.nextInt();
}
private int[] solve() {
int maxK = n * (n - 1) / 2;
if (k >= maxK) {
return new int[] {-1};
}
int[] result = new int[n];
for (int i = 0; i < n; i++) {
result[i] = i;
}
return result;
}
}
| 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);
int tot = (n * n - n) / 2;
if (tot <= K) {
printf("no solution\n");
return 0;
}
int x = 0, y = 0;
for (int i = 0; i < n; i++) {
printf("%d %d\n", 0, y);
y += 2;
}
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*(n-1)/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 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
int n, k;
cin >> n >> k;
vector<pair<int, int>> v;
int now = 1;
for (int i = 1; i <= n - 1; ++i) {
v.emplace_back(1, now);
now += 2;
}
v.emplace_back(1, now);
int tot = 0;
for (int i = 1; i <= n; ++i) {
for (int j = i + 1; j <= n; ++j) {
++tot;
}
}
if (tot > k) {
for (auto p : v) {
printf("%d %d\n", p.first, p.second);
}
} else {
cout << "no solution\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.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.StringTokenizer;
public class TheClosestPair {
public static void main(String[] args) {
MyScanner sc = new MyScanner();
int N = sc.nextInt();
int K = sc.nextInt();
if (K >= N * (N - 1) / 2) {
System.out.println("no solution");
return;
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < N; i++) {
sb.append("1 " + (9881747 - i) + "\n");
}
System.out.print(sb.toString());
}
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
int nextInt() {
return Integer.parseInt(next());
}
int[] nextIntArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
String nextLine() {
String str = "";
try { str = br.readLine(); }
catch (IOException e) { e.printStackTrace(); }
return str;
}
}
} | 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.math.BigInteger;
import java.util.StringTokenizer;
/**
* Created by Leonti on 2016-03-12.
*/
public class C {
public static void main(String[] args) {
InputReader inputReader = new InputReader(System.in);
PrintWriter printWriter = new PrintWriter(System.out, true);
int n = inputReader.nextInt();
int k = inputReader.nextInt();
int max = ((n - 1) * n) / 2;
max = max == 1 ? 1 : max == 2 ? 3 : max;
if (max <= k) {
printWriter.print("no solution");
} else {
for (int i = 0; i < n; i++) {
printWriter.println(0 + " " + i);
}
}
printWriter.close();
}
private static class InputReader {
private final BufferedReader reader;
private StringTokenizer tokenizer;
public InputReader(InputStream in) {
reader = new BufferedReader(new InputStreamReader(in));
}
public int[] nextIntArray(int size) {
int[] array = new int[size];
for (int i = 0; i < size; ++i) {
array[i] = nextInt();
}
return array;
}
public long[] nextLongArray(int size) {
long[] array = new long[size];
for (int i = 0; i < size; ++i) {
array[i] = nextLong();
}
return array;
}
public double[] nextDoubleArray(int size) {
double[] array = new double[size];
for (int i = 0; i < size; ++i) {
array[i] = nextDouble();
}
return array;
}
public String[] nextStringArray(int size) {
String[] array = new String[size];
for (int i = 0; i < size; ++i) {
array[i] = next();
}
return array;
}
public boolean[][] nextBooleanTable(int rows, int columns, char trueCharacter) {
boolean[][] table = new boolean[rows][columns];
for (int i = 0; i < rows; ++i) {
String row = next();
assert row.length() == columns;
for (int j = 0; j < columns; ++j) {
table[i][j] = (row.charAt(j) == trueCharacter);
}
}
return table;
}
public char[][] nextCharTable(int rows, int columns) {
char[][] table = new char[rows][];
for (int i = 0; i < rows; ++i) {
table[i] = next().toCharArray();
assert table[i].length == columns;
}
return table;
}
public int[][] nextIntTable(int rows, int columns) {
int[][] table = new int[rows][columns];
for (int i = 0; i < rows; i++) {
for (int j = 0; j < columns; j++) {
table[i][j] = nextInt();
}
}
return table;
}
public long[][] nextLongTable(int rows, int columns) {
long[][] table = new long[rows][columns];
for (int i = 0; i < rows; i++) {
for (int j = 0; j < columns; j++) {
table[i][j] = nextLong();
}
}
return table;
}
public double[][] nextDoubleTable(int rows, int columns) {
double[][] table = new double[rows][columns];
for (int i = 0; i < rows; i++) {
for (int j = 0; j < columns; j++) {
table[i][j] = nextDouble();
}
}
return table;
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public BigInteger nextBigInteger() {
return new BigInteger(next());
}
public boolean hasNext() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
String line = readLine();
if (line == null) {
return false;
}
tokenizer = new StringTokenizer(line);
}
return true;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
tokenizer = new StringTokenizer(readLine());
}
return tokenizer.nextToken();
}
public String readLine() {
String line;
try {
line = reader.readLine();
} catch (IOException e) {
throw new RuntimeException(e);
}
return line;
}
}
}
| 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 | def calk(n):
return n*(n-1)/2
n,k = map(int,raw_input().split())
if calk(n) > k:
for i in range(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 | import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;
public class my_class {
public static void main(String[] args) {
Scanner ololo = new Scanner(System.in);
int n = ololo.nextInt(), k = ololo.nextInt();
if((n*(n - 1) /2) <= k){
System.out.println("no solution");
}else{
for(int i = 0; i < n; i++){
System.out.println("1 " + 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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigDecimal;
import java.math.MathContext;
import java.math.RoundingMode;
import java.util.Arrays;
import java.util.Scanner;
public class Main {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringBuilder out = new StringBuilder();
String [] sp = br.readLine().split(" ");
int n = Integer.parseInt(sp[0]);
int k = Integer.parseInt(sp[1]);
int max = (n) * (n-1);
max /= 2;
if(max <= k){
out.append("no solution");
}else{
for (int i = 0; i < n; i++) {
out.append(0 + " " + i + "\n");
}
}
System.out.print(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;
int main() {
int i, j, k, l, m, n, x, y, z, r, ans = 0, mn = INT_MAX, mx = INT_MIN,
res = 0;
cin >> n >> k;
if (k * 2 >= n * (n - 1))
cout << "no solution";
else {
for (i = 0; i < n; ++i) {
cout << 0 << " " << n + 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 | #include <bits/stdc++.h>
int main() {
long long p, n;
int i = 0, j = 2;
scanf("%I64d %I64d", &n, &p);
if (((n - 1) * n) / 2 <= p)
printf("no solution\n");
else
while (n--) {
printf("%d %d\n", i, j);
j += 2;
}
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 C {
private void solve() throws IOException {
int n = ni();
int k = ni();
if (n * (n - 1) / 2 <= k) {
prln("no solution");
return;
}
for (int i = 0; i < n; ++i) {
prln(0 + " " + i);
}
}
public static void main(String ... args) throws IOException { new C().run(); }
private PrintWriter pw;
private BufferedReader br;
private StringTokenizer st;
private void run() throws IOException {
pw = new PrintWriter(System.out);
br = new BufferedReader(new InputStreamReader(System.in));
solve();
br.close();
pw.close();
}
private void pr(Object o) { pw.print(o); }
private void prln(Object o) { pw.println(o); }
private long nl() throws IOException { return Long.parseLong(nt()); }
private double nd() throws IOException { return Double.parseDouble(nt()); }
private int ni()throws IOException { return Integer.parseInt(nt()); }
private String nt() throws IOException {
if (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.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>
#pragma GCC optimize("Ofast")
using namespace std;
long long power(long long b, long long e, long long m) {
if (e == 0) return 1;
if (e % 2)
return b * power(b * b % m, (e - 1) / 2, m) % m;
else
return power(b * b % m, e / 2, m);
}
long long power(long long b, long long e) {
if (e == 0) return 1;
if (e % 2)
return b * power(b * b, (e - 1) / 2);
else
return power(b * b, e / 2);
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long n, k;
cin >> n >> k;
if (n * (n - 1) / 2 <= k)
cout << "no solution";
else {
for (auto i = 1; i <= n; i++) {
cout << i << " " << i + (n + 1) * 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 | #include <bits/stdc++.h>
using namespace std;
const int oo = 0x3f3f3f3f;
int N, K;
void Read() { cin >> N >> K; }
void Print() {
if (K >= N * (N - 1) / 2) {
cout << "no solution\n";
return;
}
for (int i = 1; i <= N; ++i) cout << "0 " << i << "\n";
}
int main() {
Read();
Print();
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 | s=input().split()
n=int(s[0])
k=int(s[1])
ans=[]
a=0
for i in range(n):
for j in range(i+1,n):
a+=1
if(a<=k):
print("no solution")
else:
for i in range(-10**9,-10**9+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 | import sys
n, k = [int(el) for el in sys.stdin.readline().split()]
if (n - 1) * n / 2 <= k:
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.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
import java.util.TreeSet;
public class ProblemA {
class Point implements Comparable<Point>{
int x;
int y;
public Point(int a, int b) {
x = a;
y = b;
}
@Override
public int compareTo(Point f) {
if(x < f.x) return -1;
else if(x > f.x) return 1;
else if(y < f.y) return -1;
return 1;
}
}
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n = s.nextInt();
long k = s.nextLong();
long limit = 1000000000;
long py[] = new long[n];
for (int i = 0; i < n; i++) {
py[i] = limit;
limit--;
}
long tot = 0;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
tot++;
}
}
if(tot > k){
for (int i = 0; i < n; i++) {
System.out.println("0 "+py[i]);
}
}
else System.out.println("no solution");
}
} | 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.InputStreamReader;
import java.io.IOException;
import java.util.InputMismatchException;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
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;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskC solver = new TaskC();
solver.solve(1, in, out);
out.close();
}
}
class TaskC {
public void solve(int testNumber, InputReader in, PrintWriter out) {
int n = in.nextInt();
int k = in.nextInt();
int time = 0;
for (int i = 1; i < n; ++i)
time += i;
if (time > k)
for (int i = 0; i < n; ++i)
out.println("0 " + i);
else
out.println("no solution");
}
}
class InputReader {
BufferedReader in;
StringTokenizer st;
public InputReader(InputStream stream) {
in = new BufferedReader(new InputStreamReader(stream));
eat("");
}
public int nextInt() {
return Integer.parseInt(next());
}
public String next() {
while (!st.hasMoreTokens())
eat(nextLine());
return st.nextToken();
}
public String nextLine() {
try {
return in.readLine();
} catch (IOException e) {
throw new InputMismatchException();
}
}
public void eat(String str) {
if (str == null) throw new InputMismatchException();
st = new StringTokenizer(str);
}
}
| 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 maxK = ((n - 1) * n) / 2;
if (maxK <= k) {
cout << "no solution" << endl;
} 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, i, j, tot = 0;
scanf("%d%d", &n, &k);
if (n * (n - 1) / 2 > k) {
for (i = 1; i <= n && tot < n; i++) {
for (j = i + 1; j <= n && tot < n; j++, tot++) {
printf("%d %d\n", i, j);
}
}
} 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.io.*;
import java.util.*;
public class TheClosestPair
{
public TheClosestPair(Scanner in)
{
int n, k;
int steps;
int i;
n = in.nextInt();
k = in.nextInt();
steps = (n-1)*(n)/2;
if (steps <= k)
System.out.printf("no solution%n");
else
{
int x = 0;
for (i=0; i < n; i++)
{
x += (n-i);
System.out.printf("%d %d%n", 0, x);
}
}
}
public static void main(String[] args)
{
new TheClosestPair(new Scanner(System.in));
}
}
| 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;
scanf("%d", &n);
scanf("%d", &k);
int tComplexity = n * (n - 1) / 2;
if (tComplexity > k) {
for (int i = 0; i < n; i++) {
printf("%d %d\n", 0, i);
}
} else
printf("%s\n", "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 | n, k = map(int,raw_input().split())
tot = n*(n-1)/2
if tot <= k:
print "no solution"
else:
for i in range(n):
print 13,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() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long t;
t = 1;
while (t--) {
long long n, k, i;
cin >> n >> k;
if (k >= (n * (n - 1)) / 2)
cout << "no solution";
else {
long long a[n], b[n];
for (i = 0; i < n; i++) a[i] = i + 1;
long long x = 0;
for (i = 0; i < n; i++) {
b[i] = n * i;
}
for (i = 0; i < n; i++) cout << a[i] << " " << b[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.util.Scanner;
public class C {
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");
in.close();
return;
}
for (int i = 0; i < n; i++) {
System.out.println(0+" "+i);
}
in.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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
scanf("%d%d", &n, &k);
if (n * (n - 1) / 2 > k) {
for (int i = 0; i < n; i++) printf("0 %d\n", i);
} else
printf("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 | #include <bits/stdc++.h>
using namespace std;
long long n, k;
int main() {
cin >> n >> k;
if (((n - 1) * n) / 2 <= k) {
cout << "no solution" << endl;
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>
int main() {
int n, k;
while (scanf("%d%d", &n, &k) != EOF) {
if (k >= n * (n - 1) / 2) {
printf("no solution\n");
} 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 java.util.*;
public class cf311a {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int k = in.nextInt();
int max = (n*n-n)/2;
if(max <= 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 | #!/usr/bin/python
n, k = map(int, raw_input().split())
if n*(n-1)/2 <= k:
print "no solution"
else:
for y in xrange(n):
print 0, y
| 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() {
long long n, k;
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;
}
}
}
| 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 = 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 | import java.io.*;
import java.util.*;
import java.util.Map.Entry;
public class C {
void run() throws IOException {
int n = ni(), k = ni();
if ((n * (n - 1)) / 2 <= k) {
pw.println("no solution");
} else {
for (int i = 1; i <= n; i++) {
pw.println(0 + " " + i);
}
}
// for (int i = 2; i <= 100; i++) {
// tr(i + " " + time(i) + " " + (i * (i - 1) / 2));
// }
}
int time(int n) {
int t = 0;
double[] x = new double[n];
double[] y = new double[n];
for (int i = 0; i < n; i++) {
x[i] = 0;
y[i] = i;
}
double d = 123456789;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
t++;
if (x[j] - x[i] >= d) {
break;
}
d = Math.min(d, Math.sqrt((x[j] - x[i]) * (x[j] - x[i]) + (y[j] - y[i]) * (y[j] - y[i])));
}
}
return t;
}
int[] na(int a_len) throws IOException {
int[] _a = new int[a_len];
for (int i = 0; i < a_len; i++)
_a[i] = ni();
return _a;
}
int[][] nm(int a_len, int a_hei) throws IOException {
int[][] _a = new int[a_len][a_hei];
for (int i = 0; i < a_len; i++)
for (int j = 0; j < a_hei; j++)
_a[i][j] = ni();
return _a;
}
String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
int ni() throws IOException {
return Integer.parseInt(next());
}
String nl() throws IOException {
return br.readLine();
}
void tr(String debug) {
if (!OJ)
pw.println(" " + debug);
}
static PrintWriter pw;
static BufferedReader br;
static StringTokenizer st;
static boolean OJ;
public static void main(String[] args) throws IOException {
long timeout = System.currentTimeMillis();
OJ = System.getProperty("ONLINE_JUDGE") != null;
pw = new PrintWriter(System.out);
br = new BufferedReader(OJ ? new InputStreamReader(System.in) : new FileReader(new File("C.txt")));
while (br.ready())
new C().run();
if (!OJ) {
pw.println();
pw.println(System.currentTimeMillis() - timeout);
}
br.close();
pw.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.awt.Point;
import java.io.*;
import java.math.BigInteger;
import java.util.*;
import java.util.Map.Entry;
import static java.lang.Math.*;
public class Solve 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 Solve(), "", 256 * (1L << 20)).start();
}
long timeBegin, timeEnd;
void time(){
timeEnd = System.currentTimeMillis();
System.err.println("Time = " + (timeEnd - timeBegin));
}
void debug(Object... objects){
if (!ONLINE_JUDGE){
for (Object o: objects){
System.err.println(o.toString());
}
}
}
int sm(int x){
if (x==1) return 0;else return 1;
}
int gcd(int a,int b){
if (b==0) return a;
else{
return gcd(b,a%b);
}
}
void psevdokod() throws IOException{
int n=readInt();
int[] x=new int[n];
int[] y=new int[n];
for (int i=0;i<n;i++){
x[i]=readInt();
y[i]=readInt();
}
//O(n*(n-1)/2)
double d=Double.MAX_VALUE;
int tot=0;
for (int i=0;i<n;i++){
for (int j=i+1;j<n;j++){
tot++;
if (x[j]-x[i]>=d){
break;
}
d=Math.min(d, Math.sqrt((x[j]-x[i])*(x[j]-x[i])+(y[j]-y[i])*(y[j]-y[i])));
}
}
out.println(tot);
}
void solve() throws IOException{
int n=readInt();
int k=readInt();
int k1=n*(n-1);
int k2=k*2;
if (k1<=k2){
out.println("no solution");
return;
}
for (int i=0;i<n;i++){
int x=n-i;
out.println(n+" "+x);
}
}
public void run(){
try{
timeBegin = System.currentTimeMillis();
init();
solve();
//psevdokod();
out.close();
time();
}catch (Exception e){
e.printStackTrace(System.err);
System.exit(-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, k;
scanf("%d %d", &n, &k);
if (n * 1LL * (n - 1) / 2 <= k)
printf("no solution\n");
else {
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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k, i;
cin >> n >> k;
if (k >= n * (n - 1) / 2) {
cout << "no solution" << endl;
return 0;
}
for (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 | n,k=map(int, input().split())
if(n*(n-1)//2 <=k):
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.PrintWriter;
import java.util.Scanner;
public class C185D2C {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
int n = in.nextInt(), k = in.nextInt();
if (k >= n*(n-1) / 2) {
out.println("no solution");
} else {
for (int i = 0; i < n; i++) {
out.printf("%d %d\n", 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 | import java.io.*;
import java.util.*;
public class C {
public static void main(String[] args){
FastScanner sc = new FastScanner();
int n = sc.nextInt();
int k = sc.nextInt();
if(((n * (n - 1)) / 2) <= k) {
System.out.println("no solution");
} else {
for(int i = 0; i < n; i++) {
System.out.println("0 " + i);
}
}
}
public static class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner(String s) {
try {
br = new BufferedReader(new FileReader(s));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
public FastScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String nextToken() {
while (st == null || !st.hasMoreElements()) try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(nextToken());
}
long nextLong() {
return Long.parseLong(nextToken());
}
double nextDouble() {
return Double.parseDouble(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;
inline long long mod(long long n, long long m = (long long)(1e9 + 7)) {
return (n % m + m) % m;
}
inline long long gcd(long long a, long long b) {
return (b == 0LL) ? a : gcd(b, a % b);
}
inline long long modPow(long long a, long long b,
long long m = (long long)(1e9 + 7)) {
long long ret = 1LL;
while (b > 0LL) {
if (b & 1) ret = mod(ret * a, m);
a = mod(a * a, m);
b >>= 1;
}
return ret;
}
inline bool leapYear(int year) {
return (year % 400 == 0) || (year % 100 != 0 && year % 4 == 0);
}
int main() {
int n, k;
while (scanf("%d %d", &n, &k) == 2) {
if (n * (n - 1) / 2 <= k)
printf("no solution\n");
else {
for (int y = (0); y < ((n)); y++) printf("0 %d\n", 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 | from Queue import * # Queue, LifoQueue, PriorityQueue
from bisect import * #bisect, insort
from datetime import *
from collections import * #deque, Counter,OrderedDict,defaultdict
import calendar
import heapq
import math
import copy
import itertools
def solver():
n,k = map(int, raw_input().split())
if k >= (n * (n -1)) / 2:
print "no solution"
return
for i in range(n):
print 0,i
if __name__ == "__main__":
solver()
| 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 | x = raw_input().split()
n = int(x[0])
k = int(x[1])
tot = n*(n-1)/2
if tot <= 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 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class B {
public static void main(String[] args) throws IOException {
PrintWriter out = new PrintWriter(System.out);
sc = new StringTokenizer(br.readLine());
int n = nxtInt();
int k = nxtInt();
if (k >= (n * (n - 1)) / 2)
out.println("no solution");
else
for (int i = 0; i < n; i++)
out.println(0 + " " + i);
br.close();
out.close();
}
static BufferedReader br = new BufferedReader(new InputStreamReader(
System.in));
static StringTokenizer sc;
static String nxtTok() throws IOException {
while (!sc.hasMoreTokens()) {
String s = br.readLine();
if (s == null)
return null;
sc = new StringTokenizer(s.trim());
}
return sc.nextToken();
}
static int nxtInt() throws IOException {
return Integer.parseInt(nxtTok());
}
static long nxtLng() throws IOException {
return Long.parseLong(nxtTok());
}
static int[] nxtIntArr(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nxtInt();
return a;
}
static char[] nxtCharArr() throws IOException {
return nxtTok().toCharArray();
}
} | 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 C {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt(), k = in.nextInt();
int sum = 0;
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
sum++;
StringBuilder sb = new StringBuilder();
if (sum > k) {
int x = 0, y = 0;
int d = 10000;
for (int i = 0; i < n; i++) {
sb.append(x + " " + y).append('\n');
y += d;
d--;
}
} else {
sb.append("no solution").append('\n');
}
System.out.print(sb);
}
}
| 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 | __author__ = 'liuchang'
import sys
n , k = map(int , sys.stdin.readline().split(' '))
if n * (n -1 ) / 2 <= k:
print "no solution"
else:
for i in range(n):
print "%d %d" % ( 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 | n, k = [int(i) for i in raw_input().strip().split()]
if k >= (n)*(n-1)/2:
print "no solution"
else:
print "0 0"
x = 0
y = 0
k = n - 1
for i in range(2, n+1):
x += 1
y += k
k -= 1
print x, y
| 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 = [int(x) for x in raw_input().split()]
if K>=N*(N-1)/2:
print 'no solution'
else:
for i in xrange(N):
print '0 %d'%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 | n,k = map(int, raw_input().split(' '))
if k>=(n-1)*n/2:
print "no solution"
else:
a=0
b=0
for i in range(n):
print a,b
b+=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 sys
input=sys.stdin.readline
from collections import defaultdict as dc
from collections import Counter
from bisect import bisect_right, bisect_left
import math
from operator import itemgetter
from heapq import heapify, heappop, heappush
from queue import PriorityQueue as pq
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 | #include <bits/stdc++.h>
using namespace std;
long long int modpow(long long int a, long long int n, long long int temp) {
long long int res = 1;
while (n > 0) {
res = (res * res) % temp;
if (n & 1) res = (res * a) % temp;
n /= 2;
}
return res;
}
long long int gcd(long long int a, long long int b) {
if (a == 0)
return (b);
else
return (gcd(b % a, a));
}
int main() {
int n, k;
cin >> n >> k;
if ((n * (n - 1)) / 2 > k) {
int x = 0, y = 0, tot = 0;
while (1) {
tot++;
cout << x << " " << y << endl;
y++;
if (tot == n) break;
}
} else {
cout << "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 | n,k = map(int,input().split())
if n*(n-1) <= k*2:
print('no solution')
else:
x = 11
for i in range(n-1):
print(1,x)
x+=3
print(1,x-5)
| 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.util.*;
import java.io.*;
public class a {
static long mod = 1000000007;
public static void main(String[] args) throws IOException
{
//Scanner input = new Scanner(new File("input.txt"));
//PrintWriter out = new PrintWriter(new File("output.txt"));
input.init(System.in);
PrintWriter out = new PrintWriter((System.out));
int n = input.nextInt(), k = input.nextInt();
if(k >= n*(n-1)/2) out.println("no solution");
else
{
int at = 0, diff = n;
for(int i = 0; i<n; i++)
{
out.println(0+" "+at);
at+=diff;
diff--;
}
}
out.close();
}
static long pow(long base, long p)
{
if(p==0) return 1;
if((p&1) == 0)
{
long sqrt = pow(base, p/2);
return (sqrt*sqrt)%mod;
}
return (base*pow(base, p-1))%mod;
}
static long modinv(long x)
{
return pow(x, mod-2);
}
/*
Sum Interval Tree - uses O(n) space
Updates and queries over a range of values in log(n) time
*/
static class IT
{
int[] left,right, a, b;
long[] val;
IT(int n)
{
left = new int[4*n];
right = new int[4*n];
val = new long[4*n];
a = new int[4*n];
b = new int[4*n];
init(0,0, n);
}
int init(int at, int l, int r)
{
a[at] = l;
b[at] = r;
if(l==r)
left[at] = right [at] = -1;
else
{
int mid = (l+r)/2;
left[at] = init(2*at+1,l,mid);
right[at] = init(2*at+2,mid+1,r);
}
return at++;
}
//return the sum over [x,y]
long get(int x, int y)
{
return go(x,y, 0);
}
long go(int x,int y, int at)
{
if(at==-1) return 0;
if(x <= a[at] && y>= b[at]) return val[at];
if(y<a[at] || x>b[at]) return 0;
return go(x, y, left[at]) + go(x, y, right[at]);
}
//add v to elements x through y
void add(int x, int y, long v)
{
go3(x, y, v, 0);
}
void go3(int x, int y, long v, int at)
{
if(at==-1) return;
if(y < a[at] || x > b[at]) return;
val[at] += (Math.min(b[at], y) - Math.max(a[at], x) + 1)*v;
go3(x, y, v, left[at]);
go3(x, y, v, right[at]);
}
}
static ArrayList<Integer>[] g;
static boolean[] marked;
static int[] id, low, stk;
static int pre, count;
static void scc()
{
id = new int[g.length]; low = new int[g.length]; stk = new int[g.length+1];
pre = count = 0;
marked = new boolean[g.length];
for(int i =0; i<g.length; i++)
if(!marked[i]) dfs(i);
}
static void dfs(int i)
{
marked[stk[++stk[0]]=i] = true;
int min = low[i] = pre++;
for(int j: g[i])
{
if(!marked[j]) dfs(j);
if(low[j] < min) min = low[j];
}
if(min < low[i]) low[i] = min;
else
{
while(stk[stk[0]] != i)
{
int j =stk[stk[0]--];
id[j] = count;
low[j] = g.length;
}
id[stk[stk[0]--]] = count++;
low[i] = g.length;
}
}
static class input {
static BufferedReader reader;
static StringTokenizer tokenizer;
/** call this method to initialize reader for InputStream */
static void init(InputStream input) {
reader = new BufferedReader(
new InputStreamReader(input) );
tokenizer = new StringTokenizer("");
}
/** get next word */
static String next() throws IOException {
while ( ! tokenizer.hasMoreTokens() ) {
//TODO add check for eof if necessary
tokenizer = new StringTokenizer(
reader.readLine() );
}
return tokenizer.nextToken();
}
static int nextInt() throws IOException {
return Integer.parseInt( next() );
}
static double nextDouble() throws IOException {
return Double.parseDouble( next() );
}
static long nextLong() throws IOException {
return Long.parseLong( next() );
}
static String nextLine() throws IOException {
return reader.readLine();
}
}
} | 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 c {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int k = sc.nextInt();
int cur=0;
for (int i=0;i<n;i++)
for (int j=i+1;j<n;j++)
cur++;
if (cur<=k)
System.out.println("no solution");
else {
int x=4;int y=0;
for (int i=0;i<n;i++){
System.out.println(x+" "+(y+n-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>
int main() {
int n, k, i;
scanf("%d %d", &n, &k);
if (k >= n * (n - 1) / 2) {
puts("no solution");
} else {
for (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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
if (n * (n - 1) / 2 > k) {
for (int i = 0; i < n; i++) {
printf("%d %d\n", 0, i);
}
} else {
cout << "no solution";
}
}
| 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.math.*;
public class Main
{
static Input in;
static Output out;
static final boolean OJ = System.getProperty("ONLINE_JUDGE") != null;
public static void main(String[] args) throws IOException
{
in = new Input(OJ ? System.in : new FileInputStream("in.txt"));
out = new Output(OJ ? System.out : new FileOutputStream("out.txt"));
solve();
out.close();
System.exit(0);
}
private static void solve() throws IOException
{
int n = in.nextInt(), k = in.nextInt();
long tot = n * (n - 1) / 2;
if (tot > k)
{
for (int i = 0; i < n; i++)
{
out.printf("%d %d\n", 0, i);
}
}
else out.println("no solution");
}
}
class Input
{
final int SIZE = 8192; //4096; 8192; 65536;
InputStream in;
byte[] buf = new byte[SIZE];
int last, current, total;
public Input(InputStream stream) throws IOException
{
in = stream; last = read();
}
int read() throws IOException
{
if (total == -1) return -1;
if (current >= total)
{
current = 0; total = in.read(buf);
if (total <= 0) return -1;
}
return buf[current++];
}
void advance() throws IOException
{
while (true)
{
if (last == -1) return;
if (!isValidChar(last)) last = read();
else break;
}
}
boolean isValidChar(int c) { return c > 32 && c < 127; }
public boolean endOfFile() throws IOException { advance(); return last == -1; }
public String nextString() throws IOException
{
advance();
if (last == -1) throw new EOFException();
StringBuilder s = new StringBuilder();
while (true)
{
s.appendCodePoint(last); last = read();
if (!isValidChar(last)) break;
}
return s.toString();
}
public String nextLine() throws IOException
{
if (last == -1) throw new EOFException();
StringBuilder s = new StringBuilder();
while (true)
{
s.appendCodePoint(last); last = read();
if (last == '\n' || last == -1) break;
}
return s.toString();
}
public String nextLine(boolean ignoreIfEmpty) throws IOException
{
if (!ignoreIfEmpty) return nextLine();
String s = nextLine();
while (s.trim().length() == 0) s = nextLine();
return s;
}
public int nextInt() throws IOException
{
advance();
if (last == -1) throw new EOFException();
int n = 0, s = 1;
if (last == '-')
{
s = -1; last = read();
if (last == -1) throw new EOFException();
}
while (true)
{
n = n * 10 + last - '0'; last = read();
if (!isValidChar(last)) break;
}
return n * s;
}
public long nextLong() throws IOException
{
advance();
if (last == -1) throw new EOFException();
int s = 1;
if (last == '-')
{
s = -1; last = read();
if (last == -1) throw new EOFException();
}
long n = 0;
while (true)
{
n = n * 10 + last - '0'; last = read();
if (!isValidChar(last)) break;
}
return n * s;
}
public BigInteger nextBigInt() throws IOException { return new BigInteger(nextString()); }
public char nextChar() throws IOException { advance(); return (char) last; }
public double nextDouble() throws IOException
{
advance();
if (last == -1) throw new EOFException();
int s = 1;
if (last == '-')
{
s = -1; last = read();
if (last == -1) throw new EOFException();
}
double n = 0;
while (true)
{
n = n * 10 + last - '0'; last = read();
if (!isValidChar(last) || last == '.') break;
}
if (last == '.')
{
last = read();
if (last == -1) throw new EOFException();
double m = 1;
while (true)
{
m = m / 10;
n = n + (last - '0') * m; last = read();
if (!isValidChar(last)) break;
}
}
return n * s;
}
public BigDecimal nextBigDecimal() throws IOException { return new BigDecimal(nextString()); }
public int[] nextIntArray(int len) throws IOException
{
int[] A = new int[len];
for (int i = 0; i < len; i++) A[i] = nextInt();
return A;
}
public long[] nextLongArray(int len) throws IOException
{
long[] A = new long[len];
for (int i = 0; i < len; i++) A[i] = nextLong();
return A;
}
public int[][] nextIntTable(int rows, int cols) throws IOException
{
int[][] T = new int[rows][];
for (int i = 0; i < rows; i++) T[i] = nextIntArray(cols);
return T;
}
}
class Output
{
final int SIZE = 4096; // 4096; 8192
Writer out;
char[] cb = new char[SIZE];
int nChars = SIZE, nextChar = 0;
char lineSeparator = '\n';
public Output(OutputStream stream) { out = new OutputStreamWriter(stream); }
void flushBuffer() throws IOException
{
if (nextChar == 0) return;
out.write(cb, 0, nextChar);
nextChar = 0;
}
void write(int c) throws IOException
{
if (nextChar >= nChars) flushBuffer();
cb[nextChar++] = (char) c;
}
void write(String s, int off, int len) throws IOException
{
int b = off, t = off + len;
while (b < t)
{
int a = nChars - nextChar, a1 = t - b;
int d = a < a1 ? a : a1;
s.getChars(b, b + d, cb, nextChar);
b += d;
nextChar += d;
if (nextChar >= nChars) flushBuffer();
}
}
void write(String s) throws IOException { write(s, 0, s.length()); }
public void print(Object obj) throws IOException { write(String.valueOf(obj)); }
public void println(Object obj) throws IOException
{
write(String.valueOf(obj));
write(lineSeparator);
}
public void printf(String format, Object... obj) throws IOException { write(String.format(format, obj)); }
public void printElements(Object... obj) throws IOException
{
for (int i = 0; i < obj.length; i++)
{
if (i > 0) print(" ");
print(obj[i]);
}
print(lineSeparator);
}
public void close() throws IOException
{
flushBuffer();
out.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.util.*;
import java.io.*;
import java.math.*;
import java.awt.geom.*;
import static java.lang.Math.*;
public class Solution implements Runnable {
long mod1 = (long) 1e9 + 7;
int mod2 = 998244353;
public void solve() throws Exception {
long n = sc.nextLong();
long k=sc.nextLong();
long max=n*(n-1)/2l;
if(k<max) {
for(int i=0;i<n;i++) {
out.println(0+" "+i);
}
}
else {
out.println("no solution");
}
}
static long gcd(long a, long b) {
if (a == 0)
return b;
return gcd(b % a, a);
}
static void sort(int[] a) {
ArrayList<Integer> l = new ArrayList<>();
for (int i : a)
l.add(i);
Collections.sort(l);
for (int i = 0; i < a.length; i++)
a[i] = l.get(i);
}
static long ncr(int n, int r, long p) {
if (r > n)
return 0l;
if (r > n - r)
r = n - r;
long C[] = new long[r + 1];
C[0] = 1;
for (int i = 1; i <= n; i++) {
for (int j = Math.min(i, r); j > 0; j--)
C[j] = (C[j] + C[j - 1]) % p;
}
return C[r] % p;
}
void sieveOfEratosthenes(boolean prime[], int size) {
for (int i = 0; i < size; i++)
prime[i] = true;
for (int p = 2; p * p < size; p++) {
if (prime[p] == true) {
for (int i = p * p; i < size; i += p)
prime[i] = false;
}
}
}
static int LowerBound(int a[], int x) { // smallest index having value >= x
int l = -1, r = a.length;
while (l + 1 < r) {
int m = (l + r) >>> 1;
if (a[m] >= x)
r = m;
else
l = m;
}
return r;
}
static int UpperBound(int a[], int x) {// biggest index having value <= x
int l = -1, r = a.length;
while (l + 1 < r) {
int m = (l + r) >>> 1;
if (a[m] <= x)
l = m;
else
r = m;
}
return l + 1;
}
public long power(long x, long y, long p) {
long res = 1;
// out.println(x+" "+y);
x = x % p;
if (x == 0)
return 0;
while (y > 0) {
if ((y & 1) == 1)
res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
static Throwable uncaught;
BufferedReader in;
FastScanner sc;
PrintWriter out;
@Override
public void run() {
try {
in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
sc = new FastScanner(in);
solve();
} catch (Throwable uncaught) {
Solution.uncaught = uncaught;
} finally {
out.close();
}
}
public static void main(String[] args) throws Throwable {
Thread thread = new Thread(null, new Solution(), "", (1 << 26));
thread.start();
thread.join();
if (Solution.uncaught != null) {
throw Solution.uncaught;
}
}
}
class FastScanner {
BufferedReader in;
StringTokenizer st;
public FastScanner(BufferedReader in) {
this.in = in;
}
public String nextToken() throws Exception {
while (st == null || !st.hasMoreTokens()) {
st = new StringTokenizer(in.readLine());
}
return st.nextToken();
}
public int nextInt() throws Exception {
return Integer.parseInt(nextToken());
}
public int[] readArray(int n) throws Exception {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public long nextLong() throws Exception {
return Long.parseLong(nextToken());
}
public double nextDouble() throws Exception {
return Double.parseDouble(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 | 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 C {
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 |
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