Nefu 118 n! How many 0 at the back (n! Power factor problem)

How many 0 in the back of n!?

problem:118 Time limit:1000ms Memory limit:65536k

Description

Read a number n from the input to find out the n! The number of the end 0.

Input

The input has several lines. The first line has an integer m, indicating the number of digits that follow. Then the M-line, each row contains a positive integer n,1<=n<=1000000000 that is determined.

Output

For each data in the input row, n, the output line, whose content is n! The number of the end 0.

Sample_input

331001024

Sample_output

024253

Ideas:

The power of p in prime factor decomposition of n! is: [n/p]+[n/p^2]+[n/p^3]+ ...

Proof in http://www.cnblogs.com/openorz/archive/2011/11/14/2248992.html detail

From the formula can be easily seen n! The power of the factor p decreases with the increase of P

For this problem, you can know that at the end of the 0 well-known factor 2*5 generation, and the power of 2 must be greater than 5, so the number of 0 is equal to (2,5) the number of matches, that is, 5 of the number of vegetarian factors

The conclusion formula can be solved by O (logn).

Accepted 820k 2ms C + + (g++ 3.4.3) 331  #include <cstdio> #include <iostream> #include <cstring># Include<algorithm>using namespace Std;int n;int main () {    int T;    scanf ("%d", &t);    while (t--)    {        scanf ("%d", &n);        int ans=0;        int p=5;        while (n/p) ans+=n/p,p*=5;        printf ("%d\n", ans);    }    return 0;}

Nefu 118 n! How many 0 at the back (n! Power factor problem)