Super Resolution

Super Resolution
Accepted : 121		Submit : 187
Time Limit : 1000 MS		Memory Limit : 65536 KB

 

Super Resolution

Bobo has an n×m picture consists of black and white pixels. He loves the picture so he would like to scale it a×b times. That is, to replace each pixel with a×b  block of pixels with the same color (see the example for clarity).

Input

The input contains zero or more test cases and is terminated by end-of-file. For each test case,

The first line contains four integers n,m,a,b . The i-th of the following n lines contains a binary string of length m which denotes the i-th row of the original picture. Character "0" stands for a white pixel while the character "1" stands for black one.

• 1≤n,m,a,b≤10
• The number of tests cases does not exceed 10.

Output

For each case, output n×a rows and m×b columns which denote the result.

Sample Input

2 2 1 1
10
11
2 2 2 2
10
11
2 2 2 3
10
11

Sample Output

10
11
1100
1100
1111
1111
111000
111000
111111
111111

 

 

/简单模拟题，将一个 n*m 矩阵放大 a,b 倍

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define MX 205

int num[MX][MX];
int ans[MX][MX];


int main()
{
    int n,m,a,b;
    while (scanf("%d%d%d%d",&n,&m,&a,&b)!=EOF)
    {
        for (int i=1;i<=n;i++)
        {
            char str[20];
            scanf("%s",str);
            for (int j=1;j<=m;j++)
                num[i][j]=str[j-1]-'0';
        }
        for (int i=1;i<=n;i++)
            for (int j=1;j<=m;j++)
                for (int x=1;x<=a;x++)
                    for (int y=1;y<=b;y++)
                        ans[(i-1)*a+x][(j-1)*b+y]=num[i][j];
        for (int i=1;i<=n*a;i++)
        {
            for (int j=1;j<=m*b;j++)
                printf("%d",ans[i][j]);
            putchar(10);
        }
    }
    return 0;
}