题目描述

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样例

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算法1

(暴力枚举) $O(n^2)$

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时间复杂度

参考文献

C++ 代码

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算法2

(暴力枚举) $O(n^2)$

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时间复杂度

参考文献

C++ 代码

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#include<bits/stdc++.h>
using namespace std;

unordered_map<string, int> months {{"January",1}, {"February", 2}, {"March", 3},
                                                {"April",4}, {"May", 5}, {"June", 6},
                                                {"July",7}, {"August", 8}, {"September", 9},
                                                {"October",10}, {"November", 11}, {"December", 12}};
unordered_map<int, int> months_day {{1,31}, {2, 28}, {3, 31},
                                    {4,30}, {5, 31}, {6, 30},
                                    {7,31}, {8, 31}, {9, 30},
                                    {10,31}, {11, 30}, {12, 31}};
unordered_map<int,string> week_day {{1,"Monday"},{2,"Tuesday"},{3,"Wednesday"},
                                    {4,"Thursday"},{5,"Friday"},{6,"Saturday"},{0,"Sunday"},};
int main()
{
    int day,year;
    string month;

    int y[3010],m[12];
    y[0] = 0,m[0] = 0;

    //初始化
    for(int i = 1;i <= 3000;i ++)
        if(((i % 4 == 0)&&(i % 100 != 0))||(i % 400 == 0))y[i] = y[i-1] + 366;
        else y[i] = y[i-1] + 365;

    for(int i = 1;i <= 12;i ++)
        m[i] = m[i-1] + months_day[i];

    while(cin >> day)
    {
        cin >> month >> year;
        int day_nums = 0;
        day_nums += y[year - 1] + m[months[month] - 1] + day;
        if((((year % 4 == 0)&&(year % 100 != 0))||(year % 400 == 0))&&(months[month] > 2))day_nums += 1;
        cout << week_day[day_nums % 7] << endl;
    }


    return 0;
}