Hack It

Hack It

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 373    Accepted Submission(s): 104
Special Judge

Problem Description

Tonyfang is a clever student. The teacher is teaching he and other students "bao'sou".
The teacher drew an n*n matrix with zero or one filled in every grid, he wanted to judge if there is a rectangle with 1 filled in each of 4 corners.
He wrote the following pseudocode and claim it runs in O(n2):

let count be a 2d array filled with 0s
iterate through all 1s in the matrix:
  suppose this 1 lies in grid(x,y)
  iterate every row r:
    if grid(r,y)=1:
      ++count[min(r,x)][max(r,x)]
      if count[min(r,x)][max(r,x)]>1:
        claim there is a rectangle satisfying the condition
claim there isn't any rectangle satisfying the condition

As a clever student, Tonyfang found the complexity is obviously wrong. But he is too lazy to generate datas, so now it's your turn.
Please hack the above code with an n*n matrix filled with zero or one without any rectangle with 1 filled in all 4 corners.
Your constructed matrix should satisfy 1≤n≤2000 and number of 1s not less than 85000.

 

Input

Nothing.

 

Output

The first line should be one positive integer n where 1≤n≤2000.

n lines following, each line contains only a string of length n consisted of zero and one.

 

Sample Input

(nothing here)

 

Sample Output

3

010

000

000 (obviously it's not a correct output, it's just used for showing output format)

 

Source

2018 Multi-University Training Contest 2

 

这题是真的强,不晓得大佬是怎么想出来的.

#include <iostream>
#include <cstring>
#include <stdio.h>
#include <algorithm>
using namespace std;
int ans[3000][3000];
int n=47;
int main(){
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      int tmp=j;
      for(int k=0;k<n;k++){
        if(k!=0)tmp+=i;
        tmp%=n;
        ans[k*n+tmp][i*n+j]=1;
      }
    }
  }
  cout<<2000<<endl;
  for(int i=0;i<2000;i++){
      for(int j=0;j<2000;j++){
        cout<<ans[i][j];
      }
      cout<<endl;
  }
  return 0;
}