Hack It
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 495 Accepted Submission(s): 148
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)想了好久的构造题，赛后听题解终于听明白了orz
把2000*2000分成几个块来搞，具体分成多少个呢，有说43有说47的，又听说43不行，就用47，一定要用个素数，47*47=2209>2000,所以将2000*2000的矩阵分成47行，47列
然后以25为例，将25*25的大矩阵分为5*5的小矩阵（这里只列出前5行前25列，也就是分块后的矩阵中的第一行的全部）
10000 10000 10000 10000 10000
10000 01000 00100 00010 00001
10000 00100 00001 01000 00010
10000 00010 01000 00001 00100
10000 00001 00010 00100 01000
分别对上一行的1的位置+1取模，就可以得到了
也不知道理解的是不是对的==
#include<iostream>
using namespace std;
const int n=47;
const int maxn=3010;
int map[maxn][maxn];
int main()
{
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
for(int k=0;k<n;k++)
{
map[i*n+j][k*n+(j*k+i)%n]=1;
}
}
}
cout<<2000<<endl;
for(int i=0;i<2000;i++)
{
for(int j=0;j<2000;j++)
{
cout<<map[i][j];
}
cout<<endl;
}
return 0;
}