hihoCoder2月29日(字符串模拟)

样例输入

4
January 12, 2012
March 19, 2012
August 12, 2899
August 12, 2901
August 12, 2000
August 12, 2005
February 29, 2004
February 29, 2012

样例输出

Case #1: 1
Case #2: 0
Case #3: 1
Case #4: 3

本题涉及到求两个时段之间闰年的公式

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
bool isLeap(int y)
{
    if((y%4==0&&y%100!=0)||y%400==0)
        return true;
    return false;
}
struct Date{
    char mon[20];
    int day;
    long long year;
    Date()
    {
        memset(mon,0,sizeof(mon));
    }
};
Date Beg;
Date End;
LL Leaps(LL y)
{
    return y/4-y/100+y/400;
}
int main()
{
    int T;
    scanf("%d",&T);
    int cas=1;
    while(cas<=T)
    {
        scanf("%s",Beg.mon);
        scanf("%d,",&Beg.day);
        scanf("%lld",&Beg.year);
        scanf("%s",End.mon);
        scanf("%d,",&End.day);
        scanf("%lld",&End.year);
        LL res=0;
        if(isLeap(Beg.year))
        {
            if(!strcmp("January",Beg.mon)||!strcmp("February",Beg.mon))
                res++;
        }
        if(isLeap(End.year))
        {
            if(!strcmp("February",End.mon)&&End.day==29||(strcmp("January",End.mon)!=0&&strcmp("February",End.mon)!=0))
            {
                res++;
            }
        }
        if(End.year==Beg.year&&isLeap(Beg.year)&&isLeap(End.year))    
            res--;
        if(End.year-1>Beg.year)
            res+=(Leaps(End.year-1)-Leaps(Beg.year));
        printf("Case #%d: %lld\n",cas++,res);
    }
    
    return 0;
}

 

hihoCoder2月29日(字符串模拟)

原文：http://www.cnblogs.com/program-ccc/p/5215731.html