牛客网--day of week

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.
输入描述:
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.
输出描述:
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.

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
示例1
输入

9 October 2001
14 October 2001

输出

Tuesday
Sunday

//注意二维字符数组的初始化。
//隐藏条件就是1年1月1日是星期一，把这个时间点设为锚点

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;

char  mon[13][20]={" ","January","February","March","April","May","June","July","August","September","October","November","December"};
char  week[7][20]={"Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"};
int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

bool leap(int y){
    if(y%400==0||y%4==0&&y%100!=0) return true;
    else return false;
}
int days(int y,int m,int d){
    int sum=0;
    for(int i=1;i<y;i++){
        if(leap(i)) sum+=366;
        else sum+=365;
    }
    if(leap(y)) day[2]+=1;
    for(int j=1;j<m;j++){
        sum+=day[j];
    }
    sum+=d;
    return sum;
}

int main(){
    int y,m,d;
    char month[13];
    
    while(cin>>d>>month>>y){
        for(int i=1;i<13;i++){
            if(strcmp(month,mon[i])==0){
                m=i;
                break;
            }
        }
        int cnt1=days(y,m,d);
   
        cout<<week[cnt1%7];
    }
    return 0;
}