http://lightoj.com/volume_showproblem.php?problem=1414 1414 - February 29
PDF (English) Statistics Forum
Time Limit: 1 second(s) Memory Limit: 32 MB
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
Note
The names of the months are {"January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November" and "December"}. And the numbers of days for the months are {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30 and 31} respectively in a non-leap year. In a leap year, number of days for February is 29 days; others are same as shown in previous
line.
PROBLEM SETTER: MD. ARIFUZZAMAN ARIF
SPECIAL THANKS: JANE ALAM JAN
题意：
给你初始年月日和结束年月日，统计2.29的个数
分析：
本来以为是模拟，写了写，打算中间除以个4就能搞定，但是忘了百年不润，然后就一直WA，尽然是容斥原理（但觉得不是容斥的样子），确实巧妙
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--；
#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;
}
bool judge(int y)
{
if((y%400==0)||(y%100!=0&&y%4==0))
return 1;
return 0;
}
int main()
{
init();
int T,cas=0;
int t1,t2,y1,y2;
int m1,m2;
string s1,s2;
scanf("%d",&T);
while(T--)
{
cas++;
cin>>s1;scanf("%d,%d",&t1,&y1);
cin>>s2;scanf("%d,%d",&t2,&y2);
if(y1>y2)
{
swap(y1,y2);
swap(t1,t2);
swap(s1,s2);
}
m1=ma[s1];
m2=ma[s2];
if(m1>=3)
y1++;
if((m2<2) ||(m2==2 && t2<29))
y2--;
y1--;
int tem1=y1/4-y1/100+y1/400;
int tem2=y2/4-y2/100+y2/400;
printf("Case %d: ",cas);
printf("%d\n",tem2-tem1);
}
return 0;
}
