日期类问题

日期类问题

@(算法)

日期类问题中最基本的问题——求两个日期间的天数差。
解决这类区间问题有一个统一的思想——把原区间问题统一到起点确定的区间问题上去。

日期类问题有一个特别需要注意的要点——闰年
闰年的判断规则：当年数不能被100整除时若能被4整除，或者其能被400整除时也是闰年。
用逻辑表达式为： Year%100!=0 && Year%4==0 || Year%400==0

---

例题

题目描述

We now use the Gregorian style of dating in Russia. The leap years are years with number divisible by 4 but not divisible by 100, or divisible by 400.For example, years 2004, 2180 and 2400 are leap. Years 2004, 2181 and 2300 are not leap.
Your task is to write a program which will compute the day of week corresponding to a given date in the nearest past or in the future using today’s agreement about dating.

输入

There is one single line contains the day number d, month name M and year number y(1000≤y≤3000). The month name is the corresponding English name starting from the capital letter.

输出

Output a single line with the English name of the day of week corresponding to the date, starting from the capital letter. All other letters must be in lower case.

样例输入

9 October 2001
14 October 2001

样例输入

Tuesday
Sunday

代码块

// task.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <iostream>
#include <cstdio>
#include <cstdlib>
#define ISYEAR(x) x%100!=0 && x%4==0 || x%400==0?1:0
//判断是否是闰年
using namespace std;

//每月的天数
int dayOfMonth[13][2] = {
    0,0,
    31,31,
    28,29,
    31,31,
    30,30,
    31,31,
    30,30,
    31,31,
    31,31,
    30,30,
    31,31,
    30,30,
    31,31
};

struct Date {
    int Year;
    int Month;
    int Day;
    void nextDay() {    //计算下一天的日期
        Day++;
        if (Day > dayOfMonth[Month][ISYEAR(Year)]) {
            Day = 1;
            Month++;
            if (Month > 12) {
                Year++;
                Month = 1;
            }

        }
    }
};

//每个月的名称
char monthName[13][20] = {
    "",
    "January",
    "February",
    "March",
    "April",
    "May",
    "June",
    "July",
    "August",
    "September",
    "October",
    "November",
    "December"
};

//每天的名称
char weekName[7][20] = {
    "Sunday",
    "Monday",
    "Tuesday",
    "Wednesday",
    "Thursday",
    "Friday",
    "Saturday"
};

int buf[3001][13][32];
int main()
{
    Date date;
    int t = 0;
    date.Year = 0;
    date.Month = 1;
    date.Day = 1;
    while (date.Year != 3001) {
        buf[date.Year][date.Month][date.Day] = t;
        date.nextDay();
        t++;
    }

    int d, m, y;
    char s[20];
    while (scanf("%d%s%d", &d, s, &y) != EOF)
    {
        for (m = 1; m <= 12; m++) {
            if (strcmp(s, monthName[m])==0) {//将输入字符串与月名比较得出月数
                break;
            }
        }
        int days = buf[y][m][d] - buf[2019][2][26];
        days += 2;                    //今天是2019.02.26，星期二，故 days+2
        puts(weekName[(days % 7 + 7) % 7]);    //用7对其取模，并且保证其为非负数
    }
    return 0;
}

总结

1.善用结构体
2.判断闰年:#define ISYEAR(x) x%100!=0 && x%4==0 || x%400==0?1:0
3.循环nextDay()，将日期转换成int整型，把原区间问题统一到起点确定的区间问题上去。

        int days = buf[y][m][d] - buf[2019][2][26];
        days += 2;                  //今天是2019.02.26，星期二，故 days+2
        puts(weekName[(days % 7 + 7) % 7]); //用7对其取模，并且保证其为非负数

关键代码！ int days = buf[y][m][d] - buf[2019][2][26]可能为负数；
days % 7 + 7) % 7用7对其取模，保证其为非负数。

参考资料:计算机考研——机试指南[电子工业出版社]