> #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;
> 		}
> 	}
> }

• 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