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  • lightoj

    Time Limit: 1000MS   Memory Limit: 32768KB   64bit IO Format: %lld & %llu

    Status

    Description

    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

    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

    Sample Output

    Case 1: 1

    Case 2: 0

    Case 3: 1

    Case 4: 3

    Source

    Problem Setter: Md. Arifuzzaman Arif
    Special Thanks: Jane Alam Jan
     
    #include <map> 
    #include <cstdio>
    #include <iostream>
    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%4==0 && y%100 !=0) || y%400 ==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 a[25];
            int m, y;
            scanf("%s%d,%d", a, &m, &y);
            int t1=0, t2 =0;
            int bb=mp[a];
            if(judge(y))
            {
                if(bb>2)
                    y++;
            }
            y--;
            t1+=y/4-y/100+y/400;
            scanf("%s%d,%d", a, &m, &y);
            if(judge(y))
            {
                if(mp[a]>2|| (mp[a]== 2 && m==29))
                    y++;
            }
            y--; 
            t2= t2+y/4-y/100+y/400;
            printf("Case %d: %d
    ", Q++, t2-t1);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/soTired/p/5451026.html
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