Image Transformation

我会带着你远行 2022-08-18 02:23 156阅读 0赞

## Image Transformation ##

####  ####

##### Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^\_^ #####

## 题目描述 ##

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.

## 输入 ##

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.

## 输出 ##

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.

## 示例输入 ##

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

## 示例输出 ##

Case 1:
    2,5
    7,10
    Case 2:
    0,1,2
    3,4,3

## 提示 ##

[zoj2857][]  
有链接提示的题目请先去链接处提交程序，AC后提交到SDUTOJ中，以便查询存档。

## 来源 ##

面向对数据结构和算法不太了解的同学

## 示例程序 ##

#include<cstdlib>  
    #include<stdio.h>  
    #include<vector>  
    #include<algorithm>  
    #include<iostream>  
    using namespace std;  
    int main()  
    {  
      vector< vector<int> > red,green,blue;  
      vector<int >line;  
      int N,M,x,count=0;  
      while(cin>>N>>M,N!=0&&M!=0)  
      {  
        count++;  
          for(int i=0;i<N;i++)  
          {  
             red.push_back(line);  
             for(int j=0;j<M;j++)  
             {  
               scanf("%d",&x);  
               red[i].push_back(x);  
             }  
          }  
          for(int i=0;i<N;i++)  
          {  
             green.push_back(line);  
             for(int j=0;j<M;j++)  
             {  
               scanf("%d",&x);  
               green[i].push_back(x);  
             }  
          }  
          for(int i=0;i<N;i++)  
          {  
             blue.push_back(line);  
             for(int j=0;j<M;j++)  
             {  
               scanf("%d",&x);  
               blue[i].push_back(x);  
             }  
          }  
          printf("Case %d:\n",count);  
          for(int i=0;i<N;i++)  
          {  
           printf("%d",(red[i][0]+green[i][0]+blue[i][0])/3);  
            for(int j=1;j<M;j++)  
             {  
               printf(",%d",(red[i][j]+green[i][j]+blue[i][j])/3);  
             }  
             printf("\n");  
          }  
          red.clear();  
          green.clear();  
          blue.clear();  
      
      }  
      
       return 0;  
    }

[zoj2857]: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2857