POJ.3751时间日期格式转换【日期计算】【格式化读入】

题目链接

问题分析:
程序说明:
 标准化读入更加简洁快速!

AC程序:

#include<stdio.h>
#include<iostream>

int main() {
    int n, flag_of_ampm;
    int year, month, day;
    int hour, min, sec;
    char ampm[][3] = {"am", "pm"};
    scanf("%d", &n);
    while(n--) {
        scanf("%d/%d/%d-%d:%d:%d", &year, &month, &day, &hour, &min, &sec);

        if(hour >= 12) 
            flag_of_ampm = 1;
        else 
            flag_of_ampm = 0;

        if(hour == 0)
            hour = 12;
        else if(hour > 12)
            hour -= 12;

        printf("%02d/%02d/%04d-%02d:%02d:%02d%s\n", month, day, year, hour, min, sec, ampm[flag_of_ampm]);

    }

    system("pause");
    return 0;
}

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转载自blog.csdn.net/qq_26398495/article/details/80369068