zoj 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

#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;
int main(){
    vector<int> r, g, b;
    int n, m, test = 0, i, t;
    while(cin >> n >> m){
        if(n == 0 && m == 0)
            break;
        test++;
        r.clear();
        g.clear();
        b.clear();
        for(i = 0; i < n * m; i++){
            cin >> t;
            r.push_back(t);
        }
        for(i = 0; i < n * m; i++){
            cin >> t;
            g.push_back(t);
        }
        for(i = 0; i < n * m; i++){
            cin >> t;
            b.push_back(t);
        }
        cout << "Case " << test << ":" << endl;
        for(i = 0; i < n * m; i++){
            cout << (r[i] + g[i] + b[i]) / 3;
            if((i + 1) % m == 0)
                cout << endl;
            else
                cout << ",";
        }
    }
    //system("pause");
    return 0;
}