题解 | #日期累加#

#include <iostream>
using namespace std;

bool check_year(int year){
    if((year%4==0&&year%100!=0)||year%400==0)return true;
    return false;
}

int main() {
    int T;
    int year,month,day,n;
    int count[2][13]={{0,31,28,31,30,31,30,31,31,30,31,30,31},{0,31,29,31,30,31,30,31,31,30,31,30,31}};
    
    cin>>T;
    while (T--) { // 注意 while 处理多个 case
        cin>>year>>month>>day>>n;
        n+=day;
        while(n){
            if(month==2&&check_year(year))n-=29;
            else n-=count[0][month];
            if(n<=0){
                if(month==2&&check_year(year))n+=29;
                else n+=count[0][month];
                break;
            }
            month++;
            if(month>12){
                month=month%12;
                year+=1;
            }
        }
        printf("%04d-%02d-%02d\n",year,month,n);

    }
}
// 64 位输出请用 printf("%lld")