POJ 1401---ask N! The number of the end 0, 2 of the number must be more than 5, observed, 0 of the production is 2*5, to find this factorial line 5 of the number can be

#include <stdio.h> #include <stdlib.h>int main () {    int t,n;    while (scanf ("%d", &t)!=eof)    {        int i;        for (i=0;i<t;i++)        {            int sum=0;            scanf ("%d", &n);            while (N)            {                n/=5;                sum+=n;            }            printf ("%d\n", sum);        }    }    return 0;}

If there are 4 numbers, 5*1 5*5*5 5*5*3 5*9

1. Become 5*n1 5*n2 5*n3 5*n4 There are 4 numbers is a multiple of 5

2. Become 5*n1 5*5*n5 5*5*n6 5*n4 There are 2 numbers is a multiple of 25

3. Become 5*n1 5*5*5 5*5*n6 5*n4 There are 1 numbers is a multiple of 125

The above is to avoid repetition, only 1 layers at a time (one-off)

The number of 5 per addition is actually the number on the right

Should be the theorem of number theory: for factorial n! , the number of 1 to N is a multiple of 5 for a given N,N/5

Also for n/25,n/125

The aim is to n/5+n/25+n/125 ...

If there is a 0, the sum is

Right

 while (N) {       sum+=n/5;       N/=5;}

Interpretation: Multiple additions, and each time to see a number divided by 5,sum overlay

N/5+n/25+n/125 ... can be seen as N/5+N1/5+N2/5

N1=n/5,n2=n1/5

Loop body: Each value is a number divided by 5, and the number of each number compared to the previous count except 5

while (N) {       sum+=n/5;       n/=5; }

For a logical relationship:

A number if it is a multiple of 25, then he must be a multiple of 5, on the contrary, it is not right, so this is the pillar he can layer

POJ 1401---ask N! The number of the end 0, 2 of the number must be more than 5, observed, 0 of the production is 2*5, to find this factorial line 5 of the number can be