ZOJ Problem Set - 2857 Image Transformation

Image Transformation

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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The image stored on a computer can be represented as a matrix of pixels. In the RGB (Red-Green-Blue) color system, a pixel can be described as a triplex integer numbers. That is, the color of a pixel is in the format "r g b" where r, g and b are integers ranging from 0 to 255(inclusive) which represent the Red, Green and Blue level of that pixel.

Sometimes however, we may need a gray picture instead of a colorful one. One of the simplest way to transform a RGB picture into gray: for each pixel, we set the Red, Green and Blue level to a same value which is usually the average of the Red, Green and Blue level of that pixel (that is (r + g + b)/3, here we assume that the sum of r, g and b is always dividable by 3).

You decide to write a program to test the effectiveness of this method.

Input

The input contains multiple test cases!

Each test case begins with two integer numbers N and M (1 <= N, M <= 100) meaning the height and width of the picture, then three N * M matrices follow; respectively represent the Red, Green and Blue level of each pixel.

A line with N = 0 and M = 0 signals the end of the input, which should not be proceed.

Output

For each test case, output "Case #:" first. "#" is the number of the case, which starts from 1. Then output a matrix of N * M integers which describe the gray levels of the pixels in the resultant grayed picture. There should be N lines with M integers separated by a comma.

Sample Input

2 2
1 4
6 9
2 5
7 10
3 6
8 11
2 3
0 1 2
3 4 2
0 1 2
3 4 3
0 1 2
3 4 4
0 0

Sample Output

Case 1:
2,5
7,10
Case 2:
0,1,2
3,4,3

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Author: ZHOU, Yuan
Source: Zhejiang Provincial Programming Contest 2007

分析：

套公式的水题。矩阵的处理。

题意：

输入n，m，代表矩阵的行和列。接着输入三个矩阵，代表R,G,B三原色的矩阵，输出三个矩阵相加再求平均数得到的矩阵。

ac代码：

#include <iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
const int maxn=105;
const int maxm=105;
int a[3][maxn][maxm],b[maxn][maxm];
//g[maxn][maxm],b[maxn][maxm];
int main()
{
    int n,m,i,j,k;
    int c=0;
    while(scanf("%d%d",&n,&m)&&n&&m)
    {
        //memset(a,0,sizeof(a));
        c++;
        for(k=0;k<3;k++)
        for(i=0;i<n;i++)
        for(j=0;j<m;j++)
        {
            scanf("%d",&a[k][i][j]);
        }
        printf("Case %d:\n",c);
        for(i=0;i<n;i++)
        for(j=0;j<m;j++)
        {
             b[i][j]=(a[0][i][j]+a[1][i][j]+a[2][i][j])/3;
             if(j<m-1)
             printf("%d,",b[i][j]);
             else printf("%d\n",b[i][j]);
        }
    }
    return 0;
}