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Factorial Trailing Zeroes in C++
Here we will see how to calculate the number of trailing 0s for the result of factorial of any number. So if the n = 5, then 5! = 120. There is only one trailing 0. For 20! it will be 4 zeros as 20! = 2432902008176640000.
The easiest approach is just calculating the factorial and count the 0s. But this approach fails for a large value of n. So we will follow another approach. The trailing zeros will be there if the prime factors are 2 and 5. If we count the 2s and 5s we can get the result. For that, we will follow this rule.
Trailing 0s = Count of 5s in prime factors of factorial(n)
so Trailing 0s = $$\lvert\frac{n}{5}\rvert+\lvert\frac{n}{25}\rvert+\lvert\frac{n}{125}\rvert+...$$
To solve this, we have to follow these steps −
- set count = 0
- for i = 5, (n/i) > 1, update i = i * 5, do
- count = count + (n /i)
- return count
Example (C++)
#include <iostream>
#include <cmath>
#define MAX 20
using namespace std;
int countTrailingZeros(int n) {
int count = 0;
for (int i = 5; n / i >= 1; i *= 5)
count += n / i;
return count;
}
main() {
int n = 20;
cout << "Number of trailing zeros: " << countTrailingZeros(n);
}Input
Number of trailing zeroes: 20
Output
Number of trailing zeros: 4
Updated on: 28-Apr-2020
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