《算法笔记》3.4小节——入门模拟->日期处理 问题 E: 日期累加

把握今天，才能拥有明天！

#include <stdio.h>

bool run(int aha)
{
    if((aha%4==0&&aha%100!=0)||aha%400==0)
        return true;
    else return false;
}
 
 
int main()
{
    int day,year,month,cnt;
    int n;
    scanf("%d",&n);
    int dayofMonth[13][2] = {{0,0},{31,31},{28,29},{31,31},{30,30},
                        {31,31},{30,30},{31,31},{31,31},{30,30},{31,31},{30,30},{31,31}};
    for(int j = 0;j < n;j++){
        scanf("%d %d %d %d",&year,&month,&day,&cnt);
        for(int i = cnt;i > 0;i--){
            day++;
            if(day == dayofMonth[month][run(year)] + 1){
                month++;
                day = 1;
            }
            if(month == 13){
                year++;
                month = 1;
            }
        }
        printf("%04d-%02d-%02d\n",year,month,day);
    }
    return 0;
}