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

样例输出：

Tuesday

Sunday

提示：

Month and Week name in Input/Output:
January, February, March, April, May, June, July, August, September, October, November, December
Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday

已知年月日，计算星期的公式有两个，一个是蔡勒公式，另一个就是基姆拉尔森计算公式。

具体的我也就不写了，大家可以百度。O(∩_∩)O哈哈~

我用的是基姆拉尔森计算公式。可以试着用蔡勒公式做下。但是那个比这个麻烦一点，因为还要计算出世纪c。

//Asimple

#include <iostream>

#include <algorithm>

#include <cstring>

#include <cstdio>

#include <cctype>

#include <cstdlib>

#include <stack>

#include <cmath>

#include <set>

#include <map>

#include <string>

#include <queue>

#include <limits.h>

#define INF 0x7fffffff

using namespace std;

const int maxn = ;

typedef long long ll;

int year, days;

string moths[] = {"January", "February", "March",

                    "April", "May", "June",

                    "July", "August", "September",

                    "October", "November", "December"};

string str;

string weeks[]={"Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday","Sunday"};

int main(){

    while( cin >> days >> str >> year ){

        int m;

        for(int i=; i<; i++){

            if( str == moths[i] ){

                m = i + ;

                break;

            }

        }

        if( m ==  || m ==  ){

            m += ;

            year --;

        }

        int a = (days +  * m +  * (m + ) /  + year + year /  - year /  + year / ) % ;

        cout << weeks[a] << endl;

    }

    return ;

}

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