题解 | #日期类#

#include <iostream>
using namespace std;
int p[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int Run(int k)
{
    if((k%400==0)||((k%4==0)&&(k%100!=0)))return 1;
    else return 0;
}
int main() {
    int y,m,d;
    int n;cin>>n;
    while (cin >> y >> m >>d) { 
        if(Run(y)==1)p[2]=29;
        else p[2]=28;

        if(d==p[m]){
            d=1;
            m++;
        }
        else d++;

        if(m>12)
        {
            y++;
            m=1;
        }
        
        printf("%04d-%02d-%02d\n",y,m,d);//ymd不用取地址
    }
}