题解 | #日期累加#

#include <bits/stdc++.h>
using namespace std;

int a1[] = {0,31,28,31,30,31,30,31,31,30,31,30,31};


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

int main(){
	int cnt,y,m,d,k,sum;
	cin>>cnt;
	while(cnt--){
		cin>>y>>m>>d>>k;
		sum = d + k;
	
		for(; (sum > a1[m] && m <= 12) || m == 2; m++){
			if(isrun(y) && m == 2)
				sum -= 29;
			else 
				sum -= a1[m];
		}
		if(m > 12){
			y++;
			for(m = 1; (sum > a1[m] && m <= 12) || m == 2; m++){
				if(isrun(y) && m == 2)
					sum -= 29;
				else 
					sum -= a1[m];
			}
		}
		printf("%d-%02d-%02d\n",y,m,sum);
	}
}