How to find array[x] = array[x] + 1 - c

Question

This is my code to find facotrial of a number and to find number of occurunces of all digits in the factorial.
#include <stdio.h>
#include <stdbool.h>
int iFactorial(int iCount)
{
int iProduct = 1;
int iNumber = 1;
while (iNumber <= iCount)
{
iProduct *= iNumber;
iNumber++;
}
return iProduct;
}
int main(void)
{
int iFac[10] = {0};
int iCount = 0;
printf("Please input a Integer: ");
scanf("%d",&iCount);
iFac[iCount] = iFactorial(iCount);
printf("\nThe value of the factorial of %d is %d\n",iCount, iFac[iCount]);
int i;
int dig[10] = {0};
while (iFac <=0)
{
int n;
n= ((iFac % 10) + 1);
dig[n] = dig[n] +1;
iFac = iFac / 10;
}
for (i = 0; i > 9; i++)
{
if (dig[i+1] >0)
{
printf ("%d %d\n", i, dig[i+1]);
}
}
}
I need to find the proper method for writing array[x] = array[x] + 1

Answer 1

I think following code will clear how you wanted to count the digits.
#include<stdio.h>
int factorial (int n)
{
if ( n == 1 ) return 1 ;
return n * factorial( n-1 ) ;
}
int main()
{
//Input number
int num ;
scanf( "%d", &num ) ;
//Calculate Factorial
int fact = factorial ( num ) ;
cout<< "\nFactorial of Number is " << fact ;
//Count the frequency of Digits
int dig[10] = {0} ;
while( fact )
{
int i = fact % 10 ;
dig[i]++ ;
fact /= 10 ;
}
for ( int i = 0 ; i < 10 ; i++ )
printf("\n The digit %d is present %d times " , i , dig[i] );
return 0 ;
}
Make sure you dont input large values for calculating the factorial for a number.