2018西北工业大学夏令营第三题

每种情况单独考虑就行

#include<iostream>
using namespace std;
int main(){
	/*
	
	5
2018 07 02 08 22
2018 07 01 08 21
2018 01 01 00 00
2018 03 01 00 00
2016 03 01 00 00

	*/
	int n,y,m,d,h,f;
	cin>>n;
	int a[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
	for(int i = 0;i < n;i++){
		cin>>y>>m>>d>>h>>f;
		//需要跨越年份 
		if(m==1&&d==1&&h<16){
			printf("%d %02d %02d %02d %02d\n",y-1,12,31,24+h-16,f);
			continue;
		}
		//二月单独考虑 
		if(m==3&&d==1&&h<16){
			if((y%400==0)||(y%100!=0&&y%4==0)){
				printf("%d %02d %02d %02d %02d\n",y,2,29,24+h-16,f);
				
			}else{
				printf("%d %02d %02d %02d %02d\n",y,2,28,24+h-16,f);
				
			}
			continue; 
		}
		//需要跨月份 
		if(d==1&&h<16){
			printf("%d %02d %02d %02d %02d\n",y,m-1,a[m-1],24+h-16,f);
			continue; 
		}
		//需要跨日期 
		if(h<16){
			printf("%d %02d %02d %02d %02d\n",y,m,d-1,24+h-16,f);
			continue;
		}
		printf("%d %02d %02d %02d %02d\n",y,m,d,h-16,f);
	}
	return 0;
}