ZOJ 2857 Image Transformation （简单题）

野性酷女 2021-11-02 06:32 143阅读 0赞

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;
    }

转载于:https://www.cnblogs.com/lidaojian/archive/2012/04/01/2428969.html