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TuesdaySunday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?1:03.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]