2018 Multi-University Training Contest 2 Hack It - HDU 6313

Hack It

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 747    Accepted Submission(s): 247
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

题目大意：

让你打个n*n由0和1组成的表，其中表中任意取四个1不能组成一个矩形，且1不能少于85000个.

解题思路：

我们先找找规律，比如要打个9*9的表

则有

100 100 100

100 010 001 //根据大佬的观察，发现每3行都是有一定规律的，即为9的开方行

100 001 010 //如前三行来看 1每一行就进一位，在第一行进位的数为零开始计，如果超过3则跳回开头继续计(mod3).

010 010 010     

010 001 100

010 100 001 //接下来每一组的规律都是一样的，但每一组1的起始位置都会进一位

001 001 001

001 100 010

001 010 100

于是大佬们就推出以下公式：

for(int i=0;i<p;i++)
        for(int j=0;j<p;j++)
            for(int k=0;k<p;k++)
                mapp[j+i*p][k*p+(i+k*j)%p]=1;    //p要为质数
官方解释：取质数 p 使得 n=p^2n=p​2​​，考虑构造 p^2p​2​​ 个有比较多1的01序列使得没有任意两个有超过一个公共1        

也不知道大佬们是怎么看出规律的，被安排的明明白白。

上个代码：

​
#include<bits/stdc++.h>
#define MAXN 3333
using namespace std;
bool mapp[MAXN][MAXN];
int main()
{
	int p=47;
	for(int i=0;i<p;i++){
		for(int j=0;j<p;j++){
			for(int k=0;k<p;k++){
				mapp[j+i*p][k*p+(i+k*j)%p]=1;
			}
		}
	}
	puts("2000"); //为保证打够点，这里取2000
	for(int i=0;i<2000;i++){
		for(int j=0;j<2000;j++)
			printf("%d",mapp[i][j]);
		printf("\n");
	}
}

​