[UVA] 12439 - February 29＠Morris' Blog｜PChome Online 個人新聞台

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[UVA] 12439 - February 29

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  G

  February
  29

 


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.
InputThe first line of input will contain T (≤ 500)
denoting the number of 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.
OutputFor each case, print the case number and the number of leap
days in between two given dates (inclusive).

 
  
  Sample Input
  
  
  Output for Sample Input
  
 

  4January 12, 2012March 19, 2012August 12, 2899August 12, 2901August 12, 2000August 12, 2005February 29, 2004February 29, 2012
  Case 1: 1Case 2: 0Case 3: 1Case 4: 3
 
NoteThe 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.


#include <stdio.h>
#include <stdlib.h>
#include <string.h>

const char month[12][20] = {"January", "February", "March", "April", "May",
    "June", "July", "August", "September", "October", "November", "December"};
int count(char str[], int d, int y, bool b) {
    if(y%4 == 0 && y%100 != 0 || y%400 == 0) {
        int i, m;
        for(i = 0; i < 12; i++) {
            if(!strcmp(str, month[i])) {
                m = i+1;
                break;
            }
        }
        if(m > 2) {
            y++;
        } else if(m == 2 && d == 29) {
            if(b == true)
                y++;
            else
                y--;
        } else {
            y--;
        }
    }
    return y/4 - y/100 + y/400;
}
int main() {
    int t, d, y, Case = 0;
    int cnt1, cnt2;
    char m[50];
    scanf("%d", &t);
    while(t--) {
        scanf("%s %d, %d", m, &d, &y);
        cnt1 = count(m, d, y, false);
        scanf("%s %d, %d", m, &d, &y);
        cnt2 = count(m, d, y, true);
        printf("Case %d: %d\n", ++Case, abs(cnt2 - cnt1));
    }
    return 0;
}

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#12439#February 29

台長： Morris

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