PAT B1066图像过滤

输入样例：

3 5 100 150 0
3 189 254 101 119
150 233 151 99 100
88 123 149 0 255

 

输出样例：

003 189 254 000 000
000 233 151 099 000
088 000 000 000 255

解题思路：1、将像素点用二维数组存放起来

　　　　　2、输入的过程中进行判断，如果在替换的范围内就存储替换的数

　　　　　3、遍历二维数组，按照对应的格式输出

#include<iostream>

using namespace std;

int pp [501] [501];//图片像素点的值

int main () {
    int M, N, start_rep, end_rep, rep;//M行N列数据，替换开始点为start,结束点为end,替换数为rep
    cin >> M >> N >> start_rep >> end_rep >> rep;
    //输入数据并进行判断和替换
    for ( int i = 0; i < M; i++ ) {
        for ( int j = 0; j < N; j++ ) {
            cin >> pp [i] [j];
            if ( pp [i] [j] >= start_rep&& pp [i] [j] <= end_rep ) {
                //在替换的范围内
                pp [i] [j] = rep;
            }
        }
    }
    //按要求输出
    for ( int i = 0; i < M; i++ ) {
        for ( int j = 0; j < N; j++ ) {
            if ( j != 0 ) printf ( " " );
            if ( pp [i] [j] < 10) {
                printf ( "00%d", pp [i] [j] );
            }
            else if ( pp [i] [j] >= 10 && pp [i] [j] < 100 ) {
                printf ( "0%d", pp [i] [j] );
            }
            else {
                printf ( "%d", pp [i] [j] );
            }
        }
        printf ( "
" );
    }
    system ( "pause" );
    return 0;
}