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 = , i, t;
     while(cin >> n >> m){
         if(n ==  && m == )
             break;
         test++;
         r.clear();
         g.clear();
         b.clear();
         for(i = ; i < n * m; i++){
             cin >> t;
             r.push_back(t);
         }
         for(i = ; i < n * m; i++){
             cin >> t;
             g.push_back(t);
         }
         for(i = ; i < n * m; i++){
             cin >> t;
             b.push_back(t);
         }
         cout << "Case " << test << ":" << endl;
         for(i = ; i < n * m; i++){
             cout << (r[i] + g[i] + b[i]) / ;
             if((i + ) % m == )
                 cout << endl;
             else
                 cout << ",";
         }
     }
     //system("pause");
     return ;
 }