題目鏈接:http://acm.hdu.edu.cn/showproblem.php?pid=6313
Hack It
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 982 Accepted Submission(s): 339
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
2000*2000看成某個質數的四次方 也就是47
爲什麼是四次方 因爲可以把整個矩陣看成47*47個47*47的小方陣
小方陣裏就是通常加法意義下模47的循環加羣
如果模的是合數循環羣表上會有某幾行是重複的結果 數論書上有。。
枚舉加值和起始位置就可構造出47*47個小方陣
聽視頻講解之後反應過來然後反思自己爲什麼這麼菜。。
理解是個循環加羣之後就很好寫代碼了。。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <stack>
#include <cstring>
#include <queue>
#include <algorithm>
#define max_ 47
#define inf 0x3f3f3f3f
#define mod 1000000007
using namespace std;
int main()
{
//freopen("data.txt","w",stdout);
cout <<"2000" <<endl;
for(int i=1;i<=47;i++)//cong i kai shi
{
for(int j=1;j<=47;j++)//bu chang
{
int mid=i;
int cnt=0;
for(int k=1;k<=47;k++)
{
mid=(mid+j)%47;
for(int l=0;l<47;l++)
{
if(l==mid)
cout << "1";
else
cout <<"0";
cnt++;
if(cnt==2000)
break;
}
if(cnt==2000)
break;
}
cout << endl;
}
cout <<endl;
}
}