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

Case 1: 1

Case 2: 0

Case 3: 1

Case 4: 3

方法一：（耗时长，超时，但思想中规中矩）

1、定义月份数组，写一个查找月份的函数。

2、每个年份分别判断是否是闰年

#include<iostream>
#include<cstdio>
#include<string.h>
#include<algorithm>
using namespace std;
char month[][20]={"January", "February", "March", "April", "May", "June","July", "August", "September", "October", "November", "December"};
int find_month(char m[]){for(int i=0;i<12;i++){if(strcmp(m,month[i])==0)return i+1;}return 0;
}
bool leap(int year){if((year%4==0&&year%100!=0)||year%400==0) return true;return false;
}
int main(){int T,day1,year1,day2,year2,kase=0;char temp1[20],temp2[20];scanf("%d",&T);while(T--){int cnt=0;scanf("%s %d, %d",temp1,&day1,&year1);scanf("%s %d, %d",temp2,&day2,&year2);int month1=find_month(temp1),month2=find_month(temp2);//printf("%d %d %d %d %d %d\n",year1,month1,day1,year2,month2,day2);if(leap(year1)){if(month1<=2) cnt++;}for(int i=year1+1;i<year2;i++){//printf("%d\n",leap(i));if(leap(i)) cnt++;}if(year2!=year1&&leap(year2)){if(month2>2) cnt++;else if(month2==2&&day2==29) cnt++;}printf("Case %d: %d\n",++kase,cnt);}return 0;
}

方法二：

方法一就是最原始的思想，第一个年份判断一下，中间每个年份判断一下，最后一个年份判断一下。但是提交上去超时。

1、对方法一中的第一点，查找月份，我们可以不用strcmp函数，只用判断每个月份前两个字母就行。这函数费时。

2、对方法一中的第二点，查找总共的闰年数，我们可以不用每个年份单独判断。这里介绍一个方法cnt=(year2/4-year2/100+year2/400)-(year1/4-year1/100+year1/400);意思就是查找年份为year1+1,year1+2,...,year2的闰年的个数。比如year1=2000，year2=2004，那么cnt=1。

对这两点优化之后，代码符合要求。如下：

#include<iostream>
#include<cstdio>
#include<string.h>
#include<algorithm>
using namespace std;
bool leap(int year){if((year%4==0&&year%100!=0)||year%400==0) return true;return false;
}
int main(){int T,day1,year1,day2,year2,kase=0;char month1[20],month2[20];scanf("%d",&T);while(T--){scanf("%s %d, %d",month1,&day1,&year1);scanf("%s %d, %d",month2,&day2,&year2);int cnt=(year2/4-year2/100+year2/400)-(year1/4-year1/100+year1/400);if(leap(year1)){if(month1[0]=='J'&&month1[1]=='a'||month1[0]=='F') cnt++;}if(leap(year2)){if(month2[0]=='J'&&month2[1]=='a'||month2[0]=='F'&&day2<29) cnt--;}printf("Case %d: %d\n",++kase,cnt);}return 0;
}