A factorial of the number of end zeros

   // Given a positive integer n, n Calculate the factorial n! The end of the number "0" is contained.

• 5! = 120, the end of which the number contained in the "0" is 1;
• 10! = 3628800, which contained the end of the number "0" is 2;
• 20! = 2432902008176640000, which contained the end of the number "0" is 4.
• //analysis:
• The number of mantissa multiplying two numbers 0 is actually dependent on the number of factors 2 and 5
• Every two consecutive numbers will have a factor of 2
• Only you need to be concerned about the number of factor 5
• n = 5*K + r（0<= r <= 4）
• n! = (5*K) * (5*(K-1)) * (5*(K-2)) * ... * 5 * A
• f (n!) = g (n!) = g (5 ^ K * K! * A) = K + g (K!) = K + f (K!), where K = n / 5 (rounded) .
• 0 <= n <= 4时，f(n!)=0
• • f(5!) = 1 + f(1!) = 1
  • f(10!) = 2 + f(2!) = 2
  • f(20!) = 4 + f(4!) = 4
  • f(100!) = 20 + f(20!) = 20 + 4 + f(4!) = 24
  • f(1000!) = 200 + f(200!) = 200 + 40 + f(40!) = 240 + 8 + f(8!) = 248 + 1 + f(1) =249

. 1 #include <the iostream>
 2  the using  namespace STD;
 . 3  
. 4  int GetN_1 ( int n-)
 . 5  {
 . 6  IF (n-< . 5 )
 . 7  {
 . 8  return  0 ;
 . 9  }
 10  the else 
. 11  {
 12 is  return (n-/ . 5 + GetN_1 (n- / 5 )); // recursive factorial find a number of zeros at the end 
13 is  }
 14  }
 15  int GetN_2 ( int n-) //Cyclic manner to find a factorial of the number of end zeros 
16  {
 . 17  int m, COUNT = 0 ;
 18 is  for ( int I = . 1 ; I <= n-; I ++ )
 . 19  {
 20 is m = I;
 21 is  the while (m% . 5 == 0 )
 22 is  {
 23 is COUNT ++ ;
 24 m / = . 5 ;
 25  }
 26 is  
27  }
 28  return COUNT;
 29  }
 30  int main ()
 31 is  {
 32 cout << GetN_1(1000) << endl;
33 cout << GetN_2(1000) << endl;
34 
35 return 0;
36 }