LightOJ 1414 February 29（闰年统计+容斥原理）

题目链接：https://vjudge.net/contest/28079#problem/R

题目大意：给你分别给你两个日期（第二个日期大于等于第一个日期），闰年的2月29日称为闰日，让你求两个日期之间的闰日数量。

解题思路：①先求闰年数量，可以根据容斥原理，得到ans=(year2/4-year1/4)-(year2/100-year1/100)+(year2/400-year1/400)

　　　　　②根据year1和year2的月份特判该年是否含闰日，注意一下上面的算式中ans包含了year2而没有包含year1所以判断时也有所不同，如果year2没有闰日ans-1,如果year1有闰日ans+1。

代码：

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 using namespace std;
 char s[][]={"","January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November","December"};
 char tmp[];

 int jg_month(void){
     for(int i=;i<=;i++){
         if(!strcmp(tmp,s[i])){
             return i;
         }
     }
 }

 bool is_leap(int year){
     if(year%==&&year%!=||year%==)
         return true;
     else
         return false;
 }

 int main(){
     int T;
     scanf("%d",&T);
     int cas=;
     while(T--){
         int day1,year1,month1,day2,year2,month2;
         scanf("%s%d,%d",tmp,&day1,&year1);
         month1=jg_month();
         scanf("%s%d,%d",tmp,&day2,&year2);
         month2=jg_month();
         //根据容斥原理
         int ans=(year2/-year1/)-(year2/-year1/)+(year2/-year1/);
         //根据year1,year2的月份进行特判
         if(month1<=&&is_leap(year1))
             ans++;
         if(is_leap(year2)&&!(month2>=||month2==&&day2==))
             ans--;
         printf("Case %d: %d\n",++cas,ans);
     }
     return ;
 }