Light 1414-February 29【容斥定理】

╰+攻爆jí腚メ 2022-08-21 13:39 11阅读 0赞

1414 - February 29

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   <td style="color:rgb(121,114,104); background:rgb(245,250,250)">Time Limit:&nbsp;<span style="color:rgb(180,95,4)">1 second(s)</span></td> 
   <td style="color:rgb(121,114,104); background:rgb(245,250,250)">Memory Limit:&nbsp;<span style="color:rgb(180,95,4)">32 MB</span></td> 
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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).

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   <td> <h1 style="color:green">Sample Input</h1> </td> 
   <td> <h1 style="color:green">Output for Sample Input</h1> </td> 
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   <td> <p>4</p> <p>January 12, 2012</p> <p>March 19, 2012</p> <p>August 12, 2899</p> <p>August 12, 2901</p> <p>August 12, 2000</p> <p>August 12, 2005</p> <p>February 29, 2004</p> <p>February 29, 2012</p> </td> 
   <td> <p>Case 1: 1</p> <p>Case 2: 0</p> <p>Case 3: 1</p> <p>Case 4: 3</p> </td> 
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# 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：

题目大意就是给出两个日期，我们要找出有多少个2月29日。我们是找出每一年之前有多少个，两个相减会得出结果，容斥定理year/400-year/100+year/4.要考虑到结束日期是不是2.29.

#include<stdio.h>
    #include<string.h>
    #include<algorithm>
    #include<map> 
    #include<string>
    using namespace std;
    map<string,int>mp;
    char month[20][25]={"January",  
     "February", "March", "April",  
      "May", "June", "July", "August",  
       "September", "October", "November",  
       "December"}; 
    bool judge(int y)
    {
    	if(y%400==0||(y%100!=0&&y%4==0))
    	return true;
    	return false;
    }   
    int main()
    {
    	for(int i=0;i<12;i++)
    	{
    		mp[month[i]]=i+1;	
    	} 
    	int t,Q=1;
    	scanf("%d",&t);
    	while(t--)
    	{
    		char ss[25];
    		int m,y;
    		int t1,t2;
    		scanf("%s%d,%d",ss,&m,&y);
    		if(judge(y))
    		{
    			if(mp[ss]>2)//第一个日期下不用考虑是不是包含2.29
    			{
    				y++;
    			}
    		}
    		y--;
    		t1=y/400-y/100+y/4;
    		scanf("%s%d,%d",ss,&m,&y);
    		if(judge(y))
    		{
    			if(mp[ss]>2||(mp[ss]==2&&m==29))
    			{
    				y++;
    			}
    		}
    		y--;
    		t2=y/400-y/100+y/4;
    		printf("Case %d: %d\n",Q++,t2-t1);
    	}
    	return 0;
    }