题解 | #日期累加#

#include<cstdio>
#include<iostream>
#include<string>
#include<map>
using namespace std;

int main()
{
	int m;
	cin>>m;
	int year,mon,day,sum;
	int mday[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
	while(m--)	{
		int totalday=0;
		scanf("%d%d%d%d",&year,&mon,&day,&sum);
		for(int i=0;i<sum;i++)
		{
		if(year%400==0||year%4==0&&year%100!=0){
			mday[2]=29;
		}	
		else
		{
			mday[2]=28;
		}
		totalday++;	
		day++;
		if(day>mday[mon])
		{
			day=1;
			++mon;
			if(mon>12){
				day=1;
				mon=1;
				year++;
			}
		}
		if(totalday==sum)
			{
				break;
			}
		}
		printf("%04d-%02d-%02d\n",year,mon,day);
		
	}
	
}

依旧是nextday策略