计算2000-1-1后任意天数的年月日及星期

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#include<iostream>
using namespace std;
bool runnian(int year)
{
if(year%4==0 && year%100!=0 || year%400==0) return 1;
else return 0;
}
int main()
{
int n;
while(cin>>n)
{
int sum1=0,sum2=0,b,d,year,month,day,weekday;
int y=2000;
int c[]= {31,28,31,30,31,30,31,31,30,31,30,31};
if(n==-1) break;
weekday=n%7;
while(sum1<=n)
{
b=sum1;
if(runnian(y)) sum1+=366;
else sum1+=365;
y++;
}
year=y-1;
///年份已算出
///下面算月份
if(runnian(year)) c[1]=29;
for(int i=0; i<12; i++)
{
d=sum2;
sum2+=c[i];
if(sum2>n-b)
///必须是>
{
month=i+1;
///月份算出
break;
}
}
if(d==0) day=n-b+1;
else day=n-b-d+1;
///天已算出
cout<<year<<"-";
if(month<10) cout<<"0"<<month<<"-";
else cout<<month<<"-";
if(day<10) cout<<"0"<<day<<" ";
else cout<<day<<" ";
switch(weekday)
{
case 0:
cout<<"Saturday";
break;
case 2:
cout<<"Monday";
break;
case 3:
cout<<"Tuesday";
break;
case 4:
cout<<"Wednesday";
break;
case 5:
cout<<"Thursday";
break;
case 6:
cout<<"Friday";
break;
case 1:
cout<<"Sunday";
break;
}
cout<<endl;
}
return 0;
}