题解 | 日期累加

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

int n;
int year,month,day,add;
bool isLeapYear(int year){
	return (year%400==0)||(year%4==0&&year%100!=0);
}
int main(){
	cin>>n;
	while(n--){
		int sum=0;
		int a[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
		cin>>year>>month>>day>>add;
		if(isLeapYear(year)){
			a[2]=29;
		}
		for(int i=1;i<month;i++){
			sum+=a[i];
		}
		sum+=day;
		sum+=add;
		int temp;
	    for(int i=year;;i++){
			if(isLeapYear(i)){
				temp=366;
			}else{
				temp=365;
			}
			if(sum<=temp){
				year=i;
				break;
			}
			sum-=temp;
		}
		int b[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
		if(isLeapYear(year)){
			b[2]=29;
		}
		int month=1;
		for(int i=1;i<=12;i++){
			if(sum<=b[i]){
				printf("%d", year);
				printf("-");
				printf("%02d", month);
				printf("-");
				printf("%02d", sum);
				break;
			}
			sum-=b[i];
			month++;
		}
		cout<<endl;
	}
		
}