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.

 1 #include <iostream>
 2 #include <cstring>
 3 #include <stdio.h>
 4 #include <algorithm>
 5 using namespace std;
 6 int ans[3000][3000];
 7 int n=47;
 8 int main(){
 9   for(int i=0;i<n;i++){
10     for(int j=0;j<n;j++){
11       int tmp=j;
12       for(int k=0;k<n;k++){
13         if(k!=0)tmp+=i;
14         tmp%=n;
15         ans[k*n+tmp][i*n+j]=1;
16       }
17     }
18   }
19   cout<<2000<<endl;
20   for(int i=0;i<2000;i++){
21       for(int j=0;j<2000;j++){
22         cout<<ans[i][j];
23       }
24       cout<<endl;
25   }
26   return 0;
27 }