February 29 容斥定理 求闰年个数

It is 2012, and it's a leap year. So there is a "February 29" in this year, which is called leap day. Interesting thing is the infant who will born in this February 29, will get his/her birthday again in 2016, which is another leap year. So February 29 only
exists in leap years. Does leap year comes in every 4 years? Years that are divisible by 4 are leap years, but years that are divisible by 100 are not leap years, unless they are divisible by 400 in which case they are leap years.

In this problem, you will be given two different date. You have to find the number of leap days in between them.

Input

Input starts with an integer T (≤ 550), denoting the number of test cases.

Each of the test cases will have two lines. First line represents the first date and second line represents the second date. Note that, the second date will not represent a date which arrives earlier than the first date. The dates will be in this format
- "month day, year", See sample input for exact format. You are guaranteed that dates will be valid and the year will be in between 2 * 103 to 2 * 109. For your convenience, the month
list and the number of days per months are given below. You can assume that all the given dates will be a valid date.

Output

For each case, print the case number and the number of leap days in between two given dates (inclusive).

Sample Input	Output for Sample Input
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

题意：给一个时间区间 求在该区间内有几个闰年
前n年有多少个闰年：num（n）=n/4-n/100+n/400

n~m年有多少闰年 ：num（m）-num（n-1）；

注意当n的月份>2月29 时，n要++；

当m的月份《2月29时，m要--；

code：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<map>
#include<string>
using namespace std;
map<string, int>ma;
void init()
{
    ma["January"]=1;
    ma["February"]=2;
    ma["March"]=3;
    ma["April"]=4;
    ma["May"]=5;
    ma["June"]=6;
    ma["July"]=7;
    ma["August"]=8;
    ma["September"]=9;
    ma["October"]=10;
    ma["November"]=11;
    ma["December"]=12;

}
int main()
{
    //freopen("i.txt","r",stdin);
    init();
    int cas,cas1=0;
    int t1,t2,y1,y2;
    int m1,m2;
    string s1,s2;
    scanf("%d",&cas);
    while(cas--)
    {
        cas1++;
        cin>>s1;scanf("%d,%d",&t1,&y1);
        
        cin>>s2;scanf("%d,%d",&t2,&y2);
        m1=ma[s1];m2=ma[s2];
        if(m1>=3)
            y1++;
        y1--;
        if((m2<2) ||(m2==2 && t2<29))
            y2--;
        int tem1=y1/4-y1/100+y1/400;
        int tem2=y2/4-y2/100+y2/400;
        
        printf("Case %d: ",cas1);
        printf("%d\n",tem2-tem1);
    }
    return 0;
}