Hack It
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 450 Accepted Submission(s): 130
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)
emmmmmmm
本来以为是找规律，题解下来后被队友仔细研读后发现是数论，可我跟数论八字不合啊~~~于是，就被带飞了
orz成则穿裙
AC代码
#include<bits/stdc++.h>
using namespace std;
const int maxn=47*47;
const int P=47;
int a[maxn+10][maxn+10];
int main()
{
int k,b,y;
cout<<"2000\n";
for(int x=0;x<maxn;x++)
{
k=x/P;
b=x%P;
for(int i=0;i<P;i++)
{
y=i*P+(k*i+b)%P;
a[x][y]=1;
}
}
long long ans=0;
for(int i=0;i<2000;i++)
{
for(int j=0;j<2000;j++)cout<<a[i][j];
cout<<'\n';
}
return 0;
}