#include<iostream.h>
#include<math.h>
int n,i,j,s;
int *m;
void main()
{
	for(;;)
	{
		cin>>n;
		if(n==0)
			break;
		for(;;)
		{
			for(i=0;;i++)
			{
				if(n/(long)pow(10,i)==0)
					break;
			}
			m=new int[i];
			for(j=0;j<i;j++)
			{
				m[j]=n/(int)pow(10,i-j-1);
				n-=m[j]*(int)pow(10,i-j-1);
			}
			s=0;
			for(j=0;j<i;j++)
				s+=m[j];
			if(s/10==0)
			{
				cout<<s<<endl;
				break;
			}
			else
				n=s;
			delete [] m;
		}
	}
}

• are you still dreaming of fitting the input into an int? (560) frkstyc 2005-04-22 00:49:48
  • 那unsigned long呢？ (0) 00 2005-04-22 00:57:57
    • use arbitrary precision arithmetic (0) frkstyc 2005-04-22 01:01:35
      • how to do it? can you help us in details? What's "arbitrary precision arithmetic"? (0) c0500301036 2005-04-22 10:19:40
        • 高精度 (0) frkstyc 2005-04-22 10:29:52
          • why your code is so short when you use "arbitrary precision arithmetic"? (10) c0500301036 2005-04-22 10:40:23
            • you may use division by 9 to find the result. also you can just add the digits up. does any of these need long code? (20) frkstyc 2005-04-22 10:56:47
              • So why you told us to use "arbitrary precision arithmetic"? I'm that confused because of this. (0) c0500301036 2005-04-22 11:03:59
                • is is, isn't it? (0) frkstyc 2005-04-22 11:15:15