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。

日期类问题有一个特别需要注意的要点——闰年

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 
    
     #include 
     
      #include 
      
       #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；

参考资料:计算机考研——机试指南[电子工业出版社]