zoj 2857

 

Image Transformation

Time Limit: 1 Second      Memory Limit: 32768 KB

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> using namespace std; int main() { int r[100][100]; int g[100][100]; int b[100][100]; int n,m; int c = 0; while(cin>>n>>m&&n!=0&&m!=0) { int i,j; for(i = 0 ; i < n; i++) for( j = 0 ; j < m; j++) cin>>r[i][j]; for(i = 0 ; i < n; i++) for( j = 0 ; j < m; j++) cin>>g[i][j]; for(i = 0 ; i < n; i++) for( j = 0 ; j < m; j++) cin>>b[i][j]; c++; printf("Case %d:/n",c); for(i = 0 ; i < n; i++) { for( j = 0 ; j < m-1; j++) cout<<(r[i][j]+g[i][j]+b[i][j])/3<<","; cout<<(r[i][m-1]+g[i][m-1]+b[i][m-1])/3<<endl; } } return 0; }