题解 第二章暴力求解| #Day of Week#

我debug了很长时间

!(Y % 4) && Y % 100 || !(Y % 400)

此处必须要在!后打括号,否则会出错

思路: 本题只需求出,今天到1000-01-01日的天数,加上一个bias

随后对7的余数

即为对应的星期

因为星期是和相差的天数息息相关的

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

static const vector<string> monthsName =
    {"January", "February", "March", "April",
     "May", "June", "July", "August", "September",
     "October", "November", "December"};

static const vector<string> weekdayName =
    {"Monday", "Tuesday", "Wednesday", "Thursday",
     "Friday", "Saturday", "Sunday"};

static const int month[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}};
static const int year[2] = {365, 366};

int main()
{
    int D = 0, M = 0, Y = 0;
    string mon;
    while (cin >> D)
    {
        cin >> mon;
        // https://blog.csdn.net/qnavy123/article/details/86977740
        // https://www.cnblogs.com/cjlaaa/p/3236727.html

        for (int i = 0; i < monthsName.size(); i++)
            if (mon == monthsName[i])
            {
                M = i + 1;
                break;
            }
        cin >> Y;
        //以上即得到了年月日

        int isRun = 0, PassDaysThisYear = 0, PassYearsDays = 0;

        if (!(Y % 4) && Y % 100 || !(Y % 400))
            isRun = 1;
        else
            isRun = 0;
        for (int i = 0; i < M; i++)
            PassDaysThisYear += month[isRun][i];
        PassDaysThisYear += D;
        //以上求出今年过去了多长时间

        for (int i = 1000; i < Y; i++)
        {
            if (!(i % 4) && (i % 100) ||!((i % 400)))
                isRun = 1;
            else
                isRun = 0;
            PassYearsDays += year[isRun];
        }
        //以上求出今年之前到1000-01-01的时间

        cout << weekdayName[(PassYearsDays + PassDaysThisYear + 1) % 7] << endl;
    }
    return 0;
}