[置顶] C语言实验:输入任意一个日期的年、月、日的值,求出从公元1年1月1日到这一天总共有多少天,并求出这一天是星期几。

要求:输入任意一个日期的年、月、日的值,求出从公元1年1月1日到这一天总共有多少天,并求出这一天是星期几。



简单的循环结构,并结合基姆拉尔森公式,注意月份转换。

下面是源码,仅供参考:

#include <iostream>

using namespace std;

void month1(int i, int count)
{
    switch (i)
    {
    case 1:count += 31;break;
    case 2:count += 29;break;
    case 3:count += 31;break;
    case 4:count += 30;break;
    case 5:count += 31;break;
    case 6:count += 30;break;
    case 7:count += 31;break;
    case 8:count += 31;break;
    case 9:count += 30;break;
    case 10:count += 31;break;
    case 11:count += 30;break;
    case 12:count += 31;break;
    }
}
void month2(int &i, int &count)
{
    switch (i)
    {
    case 1:count += 31;break;
    case 2:count += 29;break;
    case 3:count += 31;break;
    case 4:count += 30;break;
    case 5:count += 31;break;
    case 6:count += 30;break;
    case 7:count += 31;break;
    case 8:count += 31;break;
    case 9:count += 30;break;
    case 10:count += 31;break;
    case 11:count += 30;break;
    case 12:count += 31;break;
    }
}
void day1(int a)
{
    switch(a)
    {
        case 1: cout<<"Monday"<<endl;break;
        case 2: cout<<"Tuesday"<<endl;break;
        case 3: cout<<"Wednesday"<<endl;break;
        case 4: cout<<"Thursday"<<endl;break;
        case 5: cout<<"Friday"<<endl;break;
        case 6: cout<<"Saturday"<<endl;break;
        case 7: cout<<"Sunday"<<endl;break;
    }
}
int main()
{
    int year, month, day;
    while(1)
    {
        int count = 0;
        cout<<"Please input the date as the following format\nyear-month-day\n\n";
        cin>>year>>month>>day;
        for(int i = 1; i < year; i ++)
        {
            if((i % 4 == 0 && i % 100 != 0)||(i % 400 == 0))
            {
                count += 366;
            }
            else
            {
                count += 365;
            }
        }
        for(int i = 1; i < month; i++)
        {
            if((year % 4 == 0 && year % 100 != 0)||(year % 400 == 0))
            {
                month1(i,count);
            }
            else
            {
                month2(i,count);
            }
        }
        count += day;
        cout<<endl<<"Days between 1-1-1 and "<<year<<'-'<<month<<'-'<<day<<" is(are) "<<count<<" day(s)"<<endl<<endl;
        if(month==1||month==2)
        {
            month+=12;
            year--;
        }
        int d = (2+2*month+3*(month+1)/5+year+year/4-year/100+year/400)%7;
        if(d==0)
        {
            d=7;
        }
        cout<<"That day is ";
        day1(d);
        cout<<endl;
    }
}



原文地址:https://www.cnblogs.com/xinyuyuanm/p/2992072.html