ZOJ 2857 Image Transformation （简单题）

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

解题报告：这道题就是给出三个N*M的矩阵，再求平均数即可，简单题，

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAX = 105;
int a[MAX][MAX], b[MAX][MAX], c[MAX][MAX];
int ans[MAX][MAX];
int main()
{
    int n, m, i, j, count = 1;
    while (scanf("%d%d", &n, &m) != EOF)
    {
        if (n == 0 || m == 0)
        {
            break;
        }
        memset(ans, 0, sizeof(ans));
        for (i = 1; i <= n; ++i)
        {
            for (j = 1; j <= m; ++j)
            {
                scanf("%d", &a[i][j]);
            }
        }
        for (i = 1; i <= n; ++i)
        {
            for (j = 1; j <= m; ++j)
            {
                scanf("%d", &b[i][j]);
            }
        }
        for (i = 1; i <= n; ++i)
        {
            for (j = 1; j <= m; ++j)
            {
                scanf("%d", &c[i][j]);
            }
        }
        for (i = 1; i <= n; ++i)//求平均数
        {
            for (j = 1; j <= m; ++j)
            {
                ans[i][j] = (a[i][j] + b[i][j] + c[i][j]) / 3; 
            }
        }
        printf("Case %d:\n", count);
        for (i = 1; i <= n; ++i)
        {
            for (j = 1; j < m; ++j)
            {
                printf("%d,", ans[i][j]);
            }
            printf("%d\n", ans[i][m]);
        }
        count ++;
    }
    return 0;
}