UVA 10922 - 2 the 9s solution

problem link-click here

here's only one problem which is understanding how to determine the 9-degree of N. this is simple. this indicates recursively how many steps needed to get 9. let 99999 is the input N.
                          now sum= 9+9+9+9+9=45 //45 is divisible by 9
                          again sum= 4+5=9 here we get 9 by two steps
1st step adding string and 2nd step is again adding the sum value.. this will continue until we get 9 from a given string and the steps we need to get 9 is our desired 9-degree N.

an accepted code is given below:


#include<bits/stdc++.h>
using namespace std;

int value(string str)
{
    int sum=0;

    for(int i=0;i<str.length();i++)
    {
        sum=sum+(str[i]-48);
    }

    return sum;
}

int recursion(int sum)
{
         int i=1;

         if(sum%9==0)
         {
             while(sum!=9 && sum>9)
             { int r=0;
         while(sum!=0)
         {
             r=r+(sum%10);
             sum=sum/10;

         }
         sum=r;

          i++;
             }

          return i;

         }

         else
         {
             return -1;
         }


}

int main()
{
    string str;

    while(cin>>str && str!="0")
    {
        int sum=value(str);

        int i=recursion(sum);
        cout<<str;

        if(i==-1)
        {
           printf(" is not a multiple of 9.\n");
        }
        else if(str=="9")
        {
            printf(" is a multiple of 9 and has 9-degree 1.\n");
        }
        else
        {
        printf(" is a multiple of 9 and has 9-degree %d.\n",i);
        }
    }

    return 0;
}