浙江省赛problem 1009

Image Transformation

Time Limit : 2000/1000ms (Java/Other)   Memory Limit :  65536/32768K (Java/Other)

Total Submission(s) : 5   Accepted Submission(s) : 6

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  threeN *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 ofN *M integers which  describe the gray levels of the pixels in the resultant grayed picture. There  should beN lines withM 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

 

 

模拟水

题目大意：html里的RGB配色，输入点阵规模，及每点的RGB值，求中值按点阵格式输出。

题目分析：读题不难（有html基础的话）输入格式有点坑，你得分析出来它输入点阵规模后先按格式输入了R，然后是G，最后B。

#include<stdio.h>int main(){    int a[102][102],m,n,i,j,k,flag,sum=1;    while(scanf("%d%d",&m,&n)!=EOF&&m&&n)    {        for(i=0; i<m; i++)        {            for(j=0; j<n; j++)            {                a[i][j]=0;            }        }        k=3;        while(k--)        {            for(i=0; i<m; i++)            {                for(j=0; j<n; j++)                {                    scanf("%d",&flag);                    a[i][j]+=flag;                }            }        }        printf("Case %d:\n",sum);        for(i=0; i<m; i++)        {            for(j=0; j<n; j++)            {                if(j)printf(",");                printf("%d",a[i][j]/3);            }            printf("\n");        }        sum++;    }    return 0;}

本题核心算法：求三个数的平均数…………