Hard Code

Time Limit: 1 Sec  

Memory Limit: 256 MB

题目连接

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=95149#problem/A

Description

Some strange code is sent to Da Shan High School. It’s said to be the prophet’s note. The note is extremely hard to understand. However, Professor Meng is so smart that he successfully found the pattern of the code. That is, the length of the code is the product of two prime numbers. He tries to reallocate the code into a grid of size N*M, where M is the bigger prime. In specific, he writes down the letters of the code to the cells one by one, from left to right, and from top to button. In this way, he found the code eventually readable.

Professor Meng wants to know all the secrets of the note right now. But he doesn’t take his computer with him. Can you help him?

Input

The first line of the input is L (L ≤ 1000), which means the number of test cases.

For each test case, the first line contains two prime numbers, which indicates N, M (0 < N * M ≤ 1000) as describe above. The second line contains a string, i.e., the code, containing only lowercase letters. It’s guaranteed the length of the string equals to N * M. The first line of the input is L (L ≤ 1000), which means the number of test cases.

For each test case, the first line contains two prime numbers, which indicates N, M (0 < N * M ≤ 1000) as describe above. The second line contains a string, i.e., the code, containing only lowercase letters. It’s guaranteed the length of the string equals to N * M.  
C​i​​，即此题的初始分值、每分钟减少的分值、dxy做这道题需要花费的时间。

Output

For each test case, output N lines, each line with M letters representing the readable code after the reallocation.

Sample Input

1 2 5 klmbbileay

Sample Output

klmbb ileay

HINT

题意

给你一个字符串，然后变成n*m的矩阵

题解：

水题啦

代码：

#include<stdio.h>
#include<iostream>
using namespace std;

string s;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;scanf("%d%d",&n,&m);
        cin>>s;
        for(int i=0;i<s.size();i++)
        {
            cout<<s[i];
            if((i+1)%m==0)
                cout<<endl;
        }
    }
}