闰日闰年问题 大年份计算

1414 - February 29

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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.

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PROBLEM SETTER: MD. ARIFUZZAMAN ARIF

SPECIAL THANKS: JANE ALAM JAN

解题思路：

计算两个年份间闰年的个数，可以先计算两个年份分别到公元0年之间的闰年个数，然后相减，就可以得到这两个年份间的闰年个数。比如说，计算x到y之间的闰年个数，就可以先计算到公元0年的闰年个数，y/4-y/100+y/400, (x-1)/4-(x-1)/100+(X-1)/400;

然后两者相减，即可得到结果。

代码如下：

#include <iostream>#include <string>#include <cstdio>#include <fstream>using namespace std;int count(int x,int y){   int n = y/4-y/100+y/400-((x-1)/4-(x-1)/100+(x-1)/400);   return n;}int main(){    freopen("in.txt","r",stdin);    string mon[12]={"January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"};    int T,t=0;    int x,y,y1,y2,a1,a2;    char str1[100],str2[100];    scanf("%d",&T);    while(T--)    {        getchar();        scanf("%s%d, %d",str1,&a1,&y1);        scanf("%s%d, %d",str2,&a2,&y2);        int i,j;        for(i=0;i<12;i++)//判断月份            if(mon[i]==str1)                break;        for(j=0;j<12;j++)            if(mon[j]==str2)                break;        if(i+1>2)//如果开始时间大于2月29日，则直接将x设为开始时间的下一年            x=y1+1;        else            x=y1;        if(j+1>2)//如果结束时间小于2月29日，则直接将y设为结束时间的前一年            y=y2;        else if(j==1&&a2==29)            y=y2;        else            y=y2-1;        int ans=count(x,y);        printf("Case %d: %d\n",++t,ans);    }}