Solution of ZOJ 2857 Image Transformation

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Problem description

 

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

Source: Zhejiang Provincial Programming Contest 2007

此题关键在数据的读取，其余较为常规。下面给出我的实现代码：

 

#include <iostream> #include <cstdio> using namespace std; int main() { int rows, cols; int ** red_img = NULL; int ** green_img = NULL; int ** blue_img = NULL; int idx = 0; while( cin >> rows >> cols ) { if( rows == 0 || cols == 0 ) break; red_img = new int*[rows]; green_img = new int*[rows]; blue_img = new int*[rows]; for( int j = 0; j < rows; ++j ) { red_img[j] = new int[cols]; green_img[j] = new int[cols]; blue_img[j] = new int[cols]; } // read R channel for( int i = 0; i < rows; ++i ) { for( int j = 0; j < cols; ++j ) { cin >> red_img[i][j]; } } // read G channel for( int i = 0; i < rows; ++i ) { for( int j = 0; j < cols; ++j ) { cin >> green_img[i][j]; } } // read B channel for( int i = 0; i < rows; ++i ) { for( int j = 0; j < cols; ++j ) { cin >> blue_img[i][j]; } } printf( "Case %d:/n", ++idx ); int gray = 0; for( int i = 0; i < rows; ++i ) { for( int j = 0; j < cols; ++j ) { cout << ( red_img[i][j] + green_img[i][j] + blue_img[i][j] ) / 3 << ( ( j == cols - 1 ) ? "/n" : "," ); } } for( int i = 0; i < rows; ++i ) { delete [] red_img[i]; delete [] green_img[i]; delete [] blue_img[i]; } delete [] red_img; delete [] blue_img; delete [] green_img; red_img = blue_img = green_img = NULL; } return 0; }