[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。

#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;
}
}