算法笔记 — Day of Week_算法day of the week-优快云博客

本文链接：https://blog.youkuaiyun.com/m0_37345402/article/details/83620270

本文介绍了一个程序，用于计算1000年至3000年间任意给定日期是星期几。通过比较该日期与设定的初始日期（999年12月31日，星期二），程序计算出两日期之间的天数差，然后确定目标日期对应的星期。程序使用了格里高利历规则来判断闰年，并考虑了不同月份的天数。

题目链接：http://codeup.cn/problem.php?cid=100000578&pid=1

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

21 December 2012
5 January 2013

样例输出

Friday
Saturday

年的范围是1000-3000，设一个初始日期999年12月31日为周二，先求出题目所给日期距离此日期的天数，再将天数转化为星期几输出。

#include<iostream>
#include<cstring>
using namespace std;
int day_month[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}};
char day_week[8][11]={"Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"};
char month_year[13][20]={"","January","February","March","April","May",
"June","July","August","September","October","November","December"
};
bool is_leap(int year){
    if(year%400==0||(year%4==0&&year%100!=0)){
        return 1;
    }
    return 0;
}
int myfun(int y2,int m2,int d2){
	int y1=999,m1=12,d1=31;//999年12月31日是周二 
	int num1=y1*10000+m1*100+d1;
	int num2=y2*10000+m2*100+d2;
	/*if(num1>num2){
		swap(num1,num2);
	}*/
	int cnt=0;
	while(y1<y2||m1<m2||d1<d2){
		d1++;
		if(d1>day_month[m1][is_leap(y1)]){
			m1+=1;
			d1=1;
		}
		if(m1==13){
			y1++;
			m1=1;
		}
		cnt++;
	}
    return cnt;
}
int main(){
    int y,d;
    char m[13];
    int mm;
    while(cin>>d>>m>>y){
        for(int i=1;i<=12;i++){
            if(strcmp(m,month_year[i])==0){
                mm=i;
                break;
            }
        }
        int ans = myfun(y,mm,d); //ans为相差天数 
        cout<<day_week[(ans+1)%7]<<endl;
    }
    return 0;
}