11332 - Summing Digits

11332 - Summing Digits

Problem J: Summing Digits

For a positive integer n, let f(n) denote the sum of the digits of n when represented in base 10. It is easy to see that the sequence of numbers n, f(n), f(f(n)), f(f(f(n))), ... eventually becomes a single digit number that repeats forever. Let this single digit be denoted g(n).

For example, consider n = 1234567892. Then:

f(n) = 1+2+3+4+5+6+7+8+9+2 = 47
f(f(n)) = 4+7 = 11
f(f(f(n))) = 1+1 = 2

Therefore, g(1234567892) = 2.

Each line of input contains a single positive integer n at most 2,000,000,000. For each such integer, you are to output a single line containing g(n). Input is terminated by n = 0 which should not be processed.

Sample input

2
11
47
1234567892
0

Output for sample input

2
2
2
2

---

Zachary Friggstad

#include<stdio.h>

int main()

{

    char c[100];

    int i,f,fff,ff,a,bb,ccc,aa,b,cc;

    while(gets(c)!=EOF)

    {

        if(c[0]=='0')

        {

            break;

        }

        int sum=0;

    int l=strlen(c);

    for(i=0;i<l;i++)

    {

        sum=sum+c[i]-'0';

    }

    if(sum<10)

    {

        printf("%d\n",sum);

    }

    else

    {

        f=sum/10;

        ff=sum%10;

        fff=f+ff;

        if(fff<10)

    {

        printf("%d\n",fff);

    }

    else

    {

        a=fff/10;

        bb=fff%10;

        ccc=f+ff;

        if(ccc<10)

    {

        printf("%d\n",ccc);

    }

    else

    {

        aa=ccc/10;

        b=ccc%10;

        cc=aa+b;

        if(cc<10)

    {

        printf("%d\n",cc);

    }

    }

    }

    }

    }

    return 0;

}

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