Number of Trailing Zeros in a Factorial in c - c

Question

I am new to C programming
I want to find out the number of trailing zeros in a Factorial of given number
{
long int n=0,facto=1,ln=0;
int zcount=0,i=1;
printf("Enter a number:");
scanf("%ld",&n);
if(n==0)
{
facto=1;
}
else
{
for(i=1;i<=n;i++)
{
facto=facto*i;
}
}
printf("%ld",facto);
while(facto>0)
{
ln=facto%10;
facto/10;
if(ln=!0)
{
break;
}
else
{
zcount+=1;
}
}
printf("Tere are Total %d Trailing zeros in given factorial",zcount);
}
I have tried to calculate the modulo of the the number which will return the last digit of given number as the remainder and then n/10; will remove the last number.
After executing the program the output always shows number of trailing zeros as "0",The condition if(ln =! 0) always gets satisfied even if there is a zero.

Answer 1

Let us remember math (wiki):
int trailingZeroCountInFactorial(int n)
{
int result = 0;
for (int i = 5; i <= n; i *= 5)
{
result += n / i;
if(i > INT_MAX / 5) // prevent integer overflow
break;
}
return result;
}

Answer 2

This
if(ln=!0)
means ln = !0 i.e. you are assigning !0 to ln so the condition is always true, change it to
if (ln != 0)
you can use an assignment as a truth value, but if you are sure you are doing it.
To prevent accidentally doing it, turn on compiler warnings.

Answer 3

The correct syntax in
if(ln!=0)
{
// do something
}
In if ( ln=!0 ) , first !0 is evaluated then = assigns !0(which is 1) to ln, then if checks the value of ln, if the value is non-zero(in this case it is 1 due to previous assignment) then break statement is executed.
Also, this approach will not work for numbers larger than 25.
It is obvious that the power of 5 in n! will be the answer.
The following expression gives the power of a prime p in n! :-
f(n/p) + f(n/p^2) + f(n/p^3) + ..... (until f(n/p^k) return zero, where f(a) returns the greatest integer of a or the floor of a

Answer 4

package lb;
import java.util.Scanner;
public class l1 {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner s=new Scanner(System.in);
int n=s.nextInt();
int a=n/5;
if(a<5)
{
System.out.println(a);
}
else
{
int c=a/5+a;
System.out.println(c);
}
}
}

Answer 5

Try this
int trailingZeroCountInFactorial(int n)
{
int result = 0;
for (int i = 5; i <= n; i *= 5)
{
result += n / i;
}
return result;
}
To learn, how it works, there is a great explanation here

Answer 6

I have solved this kind of problem, I think your question is just find the number of trailing zeros of a factorial number like - 15! = 1307674368000 if you look at the trailing 3 digits which are 000 Efficient code is
int n;cin>>n;
int ans = 0;
while(n)
{ ans += (n = n/5);}
cout<<ans;
Here one should look for int overflow, array bounds. Also, the special cases are n=1?

Answer 7

{
long int n=0,facto=1,ln=0;
int zcount=0,i=1;
printf("Enter a number:");
scanf("%ld",&n);
if(n==0)
{
facto=1;
}
else
{
for(i=1;i<=n;i++)
{
facto=facto*i;
}
}
printf("%ld\n",facto);
while(facto>0)
{
ln=(facto%10);
facto/=10; //here you done one mistake
if(ln!=0) //another one here
{
break;
}
else
{
zcount+=1;
}
}
printf("Tere are Total %d Trailing zeros in given factorial",zcount);
}
/Run this code it will work now and mistakes you already knows i guess/