Question

#include <iostream>
#include <string>

using namespace std; 


void dayofweek(int &m, int d, int y);
int main()
{

int m, d, y, daynum; 
char ans;

do {
cout << "Enter a date as mm dd yyyy: \n";
cin >> m >> d >> y; 

dayofweek(m, d, y);

cout << m << d << y;

daynum=(d + 5 + ((26*(m+1))/10) + ((5*(y%100))/4) + ((21*(y/100))/4) )%7;

cout << daynum;


switch(daynum)
{
    case 0:
        cout << m << "/" << d << "/" << y << "is a Monday \n";  
        break;
    case 1:
        cout << m << "/" << d << "/" << y << "is a Tuesday \n"; 
        break; 
    case 2:
        cout << m << "/" << d << "/" << y << "is a Wednesday \n";
        break;
    case 3:
        cout << m << "/" << d << "/" << y << "is a Thursday \n";
        break;
    case 4:
        cout << m << "/" << d << "/" << y << "is a Friday \n";
        break;
    case 5:
        cout << m << "/" << d << "/" << y << "is a Saturday \n";
        break;
    case 6:
        cout << m << "/" << d << "/" << y << "is a Sunday \n";
        break;
}
cout<<"Do you want to continue? (y/n) \n"; 
cin >> ans; 
}while (ans == 'y' || 'Y');
return 0;
}

void dayofweek(int& m, int d, int y) 
{
    if (m==1 || m==2)
    m=m+12;

}

这编译并且工作正常，但是当我输入日期时，会出现错误的工作日。

例如，2014年2月28日是星期五，但周六出现。有什么建议？我坚持要改变什么，我也多次检查方程式。

Answer 1

在Zeller的同意实施结果中意味着：0 = Saturday, 1 = Sunday, 2 = Monday ...... 您使用的公式还需要更改dayofweek函数和更改以处理年份。我决定抛弃dayofweek函数而不是介绍dayNumber。

将此更改应用于切换语句和函数后，我得到January 1, 2000和March 1, 2000的{{3}}示例的正确结果。另外，对于2 28 2014输入，我得到Friday作为正确答案。

更改了代码：

#include <iostream>
#include <string>
#include <cmath>

using namespace std;

int dayNumber(int m, int d, int y) {
    if (m == 1 || m == 2) {
        m = m + 12;
        y = y - 1;
    }
    return (d + (int)floor((13 * (m + 1)) / 5) + y%100 + (int)floor((y%100)/ 4) + (int)floor(((int)floor(y/100))/4) + 5*(int)floor(y/100)) % 7;
}

int main() {

    int m, d, y, daynum;
    char ans;

    do {
        cout << "Enter a date as mm dd yyyy: \n";
        cin >> m >> d >> y;
        daynum = dayNumber(m, d, y);
        cout << "daynum:" << daynum << "\t";

        switch (daynum) {
            case 2:
                cout << m << "/" << d << "/" << y << " is a Monday \n";
                break;
            case 3:
                cout << m << "/" << d << "/" << y << " is a Tuesday \n";
                break;
            case 4:
                cout << m << "/" << d << "/" << y << " is a Wednesday \n";
                break;
            case 5:
                cout << m << "/" << d << "/" << y << " is a Thursday \n";
                break;
            case 6:
                cout << m << "/" << d << "/" << y << " is a Friday \n";
                break;
            case 0:
                cout << m << "/" << d << "/" << y << " is a Saturday \n";
                break;
            case 1:
                cout << m << "/" << d << "/" << y << " is a Sunday \n";
                break;
        }
        cout << "Do you want to continue? (y/n) \n";
        cin >> ans;
    } while ((ans == 'y') || (ans == 'Y'));

    return 0;
}

泽勒的同余编码，输出错误的一天

1 个答案: