C Exercises: Check whether a given number with base b is a Niven number or not

C Numbers: Exercise-38 with Solution

Write a C programming to check whether a given number with base b (2 <= b<= 10) is a Niven number or not.
From Wikipedia,
In recreational mathematics, a harshad number (or Niven number) in a given number base, is an integer that is divisible by the sum of its digits when written in that base. Harshad numbers in base n are also known as n-harshad (or n-Niven) numbers. Harshad numbers were defined by D. R. Kaprekar, a mathematician from India. The word "harshad" comes from the Sanskrit harṣa (joy) + da (give), meaning joy-giver. The term “Niven number” arose from a paper delivered by Ivan M. Niven at a conference on number theory in 1977. All integers between zero and n are n-harshad numbers.
The number 18 is a harshad number in base 10, because the sum of the digits 1 and 8 is 9 (1 + 8 = 9), and 18 is divisible by 9 (since 18/9 = 2, and 2 is a whole number).

Test Data
Input: base 10: Number 3
Output: 3 is a Niven Number
Input: base 10: Number 18
Output: 18 is a Niven Number
Input: base 10: Number 15
Output: 15 is not a Niven Number

Sample Solution:

C Code:

#include <stdio.h>
#include <string.h>

void add(char m[],char n[],int* m_len,int n_len)
{
int i,k;
int j = 0;
for(i = 0; i< *m_len&&i<n_len; i++)
    {
        k = m[i]+n[i]+j;
m[i] = k%10;
        j = k/10;
    }
for(;i< *m_len; i++)
    {
        k = m[i]+j;
m[i] = k%10;
        j = k/10;
    }
for(;i<n_len; i++)
    {
        k = n[i]+j;
m[i] = k%10;
        j = k/10;
        (*m_len)++;
    }
if(j)
    {
m[i] = j;
        (*m_len)++;
    }
}
void mult(char n[],int m, int* len)
{
int i,j,k;
    j = 0;
for(i = 0; i< *len; i++)
    {
        k = n[i]*m+j;
        j = k/10;
n[i] = k%10;
    }
for(;j;j /= 10)
    {
n[i] = j%10;
        (*len)++;
    }
}
char div(char n[],int m, int len)
{
int i,j;
j = 0;
for(i = len-1; i>= 0; i--)
    {
        j = j % m;
j *= 10;
        j += n[i];
    }
return j % m;
}
int main()
{
int base,d,i,j;
char num[4][1024];
int len[4];
printf("Input the base and the number(separated by a space)(0 to exit):\n");
scanf("%d",&base);
while(base)
    {
getchar();
scanf("%s",num[0]);
len[1] = strlen(num[0]);
        d = 0;
for(i = 0; i<len[1]; i++)
        {
num[1][len[1]-1-i] = num[0][i]-'0';
            d += num[1][len[1]-1-i];
num[0][i] = 0;
num[2][i] = 0;
        }
num[0][0] = 1;
len[0] = 1;
len[2] = 1;
for(i = 0; i<len[1]; i++)
        {
for(j = 0; j <len[0]; j++)
            {
num[3][j] = num[0][j];
            }
len[3] = len[0];
mult(num[3],num[1][i],len+3);
add(num[2],num[3],len+2,len[3]);
mult(num[0],base,len);
        }
if(div(num[2],d,len[2]))
        {
printf("It is not a Niven number.\n");
        }
else
printf("It is a Niven number.\n");
scanf("%d",&base);
    }
return 0;
}

Sample Output:

 Input the base and the number(separated by a space)(0 to exit): 10 3 0
It is a Niven number.

Flowchart:

C Programming Code Editor:

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