SDIBT 2345 (3.2.1 factorials factorial)

Description

The factorial writing of n n! represents the product of all positive integers less than or equal to N. The factorial will quickly become larger, as 13! must be stored with a 32-bit integer type, 70! Even with floating-point numbers. Your task is to find the last non-0-bit of factorial. For example, 5!=1*2*3*4*5=120 so 5! The last side of the non-0-bit is 2,7! =1*2*3*4*5*6*7=5040, so the last non-0 bits of the surface are 4.

Input

A total of one row, an integer not greater than 4,220 of N.

Output

A total of one row, output n! the last non-0 bits.

Sample Input

7

Sample Output

4

1#include <stdio.h>2 3 intMain ()4 {5     intn,date[ the],cnt=0;6     inti,sum=1;7scanf"%d",&n);8      for(I=n; i>=1; i--)9     {Tendate[i]=i; One          while(date[i]%5==0)//Culling 5 A         { -Date[i]/=5; -cnt++; the         } -     } -      for(I=n; i>=1; i--)//eliminate the same number of 2 so that No 10 will be present. -     { +         if(cnt==0) -              Break; +          while(date[i]%2==0&&cnt!=0) A         { atDate[i]/=2; -cnt--; -         } -     } -      for(I=n; i>=1; i--) -Sum= (Sum*date[i])%Ten; inprintf"%d\n", sum); -     return 0; to}

View Code

SDIBT 2345 (3.2.1 factorials factorial)