Trailing Zeros in the Product of First 2017 factorials

The following C code solves the problem:

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#include<stdio.h>

int leading(int n){
    int i, ans = 0;
    for(i = 5; i <= n; i+=5){
        int pot = 0, j = i;
        while(j % 5 == 0){
            pot++;
            j /= 5;
        }
        ans += pot * (n - i + 1);
    }
    return ans;
}

int main(){
    int n;
    scanf("%d", &n);
    printf("%d", leading(n));
    return 0;
}

Let $x=\displaystyle \prod_{i=1}^{n} i!$ , note that we can rewrite it as $x=\displaystyle \prod_{i=1}^{n} i^{n-i+1}$ . We need to find the highest $k$ such that $10^k \mid x$ . But $10=2\times 5$ , and since $2$ will appear more times than $5$ in the prime factorization of $x$ , we only need to find the highest $k$ such that $5^k \mid x$ .

So, we iterate from $i=1$ to $i=n$ , count the number of times that $5$ can divide $i$ and multiply that by $n-i+1$ . Finally, we sum all of these results and that will be the answer.