时间限制:2000ms
单点时限:1000ms
内存限制:256MB
描述
给定两个日期，计算这两个日期之间有多少个2月29日（包括起始日期）。
只有闰年有2月29日，满足以下一个条件的年份为闰年：
1. 年份能被4整除但不能被100整除
2. 年份能被400整除
输入
第一行为一个整数T，表示数据组数。
之后每组数据包含两行。每一行格式为"month day, year"，表示一个日期。month为{"January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November" , "December"}中的一个字符串。day与year为两个数字。
数据保证给定的日期合法且第一个日期早于或等于第二个日期。
输出
对于每组数据输出一行，形如"Case #X: Y"。X为数据组数，从1开始，Y为答案。
数据范围
1 ≤ T ≤ 550
小数据：
2000 ≤ year ≤ 3000
大数据：
2000 ≤ year ≤ 2×109
样例输入
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
样例输出
Case #1: 1
Case #2: 0
Case #3: 1
Case #4: 3
| #include <iostream> |
| #include <string> |
| #include <vector> |
| using namespace std; |
| |
| class LeapYear |
| { |
| private: |
| int year1; |
| int month1; |
| int day1; |
| int year2; |
| int month2; |
| int day2; |
| |
| public: |
| int judgeMonth(string month) |
| { |
| if ("January" == month) return 1; |
| else if ("February" == month) return 2; |
| else if ("March" == month) return 3; |
| else if ("April" == month) return 4; |
| else if ("May" == month) return 5; |
| else if ("June" == month) return 6; |
| else if ("July" == month) return 7; |
| else if ("August" == month) return 8; |
| else if ("September" == month) return 9; |
| else if ("October" == month) return 10; |
| else if ("November" == month) return 11; |
| else return 12; |
| } |
| |
| void init() |
| { |
| string monthStr; |
| char sig; |
| cin >> monthStr >> day1 >> sig >> year1; |
| month1 = judgeMonth(monthStr); |
| cin >> monthStr >> day2 >> sig >> year2; |
| month2 = judgeMonth(monthStr); |
| } |
| |
| int leapYearNum() |
| { |
| int lyNum = 0; |
| int divide_4 = 0, divide_400 = 0, ndivide_100 = 0; |
| if (month1 <= 1 || (month1 <= 2 && day1 <= 29)) |
| year1 -= 1; |
| if (month2 <= 1 || (month2 <= 2 && day2<29)) |
| year2 -= 1; |
| //从年份0年到开始年的闰年个数-从年分0年到结束年的闰年个数 |
| divide_400 = year2 / 400 - year1 / 400; |
| divide_4 = year2 / 4 - year1 / 4; |
| ndivide_100 = year2 / 100 - year1 / 100; |
| lyNum = divide_4 + divide_400 - ndivide_100; |
| return lyNum; |
| } |
| }; |
| |
| int main() |
| { |
| int n = 0; |
| LeapYear leapYear; |
| while (cin >> n) |
| { |
| for (int i = 1; i <= n; i++) |
| { |
| int lyNum = 0; |
| leapYear.init(); |
| lyNum = leapYear.leapYearNum(); |
| cout << "Case #" << i << ": " << lyNum << endl; |
| } |
| } |
| return 0; |
| } |
代码地址 https://github.com/iptop/hihoCoderProblemSolving/blob/master/1148/main.cpp
