【数组】1185. 一周中的第几天（简单）

【题目】
给你一个日期，请你设计一个算法来判断它是对应一周中的哪一天。输入为三个整数：day、month 和 year，分别表示日、月、年。您返回的结果必须是这几个值中的一个 {“Sunday”, “Monday”, “Tuesday”, “Wednesday”, “Thursday”, “Friday”, “Saturday”}。
【示例 1】
输入：day = 31, month = 8, year = 2019
输出：“Saturday”
【示例 2】
输入：day = 18, month = 7, year = 1999
输出：“Sunday”
【示例 3】
输入：day = 15, month = 8, year = 1993
输出：“Sunday”
【提示】
给出的日期一定是在 1971 到 2100 年之间的有效日期。
【代码】

class Solution {
public:
    int d[2][13]={
        {0,31,28,31,30,31,30,31,31,30,31,30,31},
        {0,31,29,31,30,31,30,31,31,30,31,30,31}
    };
    int y[2]={365,366};
    int isLeapYear(int year){
        if((year%4==0&&year%100!=0)||year%400==0)
            return 1;
        return 0;
    }
    string dayOfTheWeek(int day, int month, int year) {
        string week[9]={"Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"};
        int sumday=4,flag;
        for(int i=1971;i<year;i++)
            sumday+=y[isLeapYear(i)];
        flag=isLeapYear(year);
        for(int j=1;j<month;j++)
            sumday+=d[flag][j];
        sumday+=day;
        return week[sumday%7];
    }
};

【基姆拉尔森计算公式】

class Solution {
public:
    string dayOfTheWeek(int day, int month, int year) {
        if(month==1||month==2) month+=12,year--;
	int iWeek = (day+2*month+3*(month+1)/5+year+year/4-year/100+year/400)%7;                              //基姆拉尔森计算公式
       string result[]= { "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday","Sunday"};
	return result[iWeek];
    }
};