hdu 1066 Last non-zero Digit in N! 求N!的最后一位非0数 N很大

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本文介绍了一个程序设计问题：如何计算N!（N的阶乘）的末位非零数字。通过一系列数学技巧和算法实现，该程序能够高效地解决这一问题，为读者提供了详细的代码示例及解析。

Last non-zero Digit in N!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2355 Accepted Submission(s): 497

Problem Description

The expression N!, read as "N factorial," denotes the product of the first N positive integers, where N is nonnegative. So, for example,
N N!
0 1
1 1
2 2
3 6
4 24
5 120
10 3628800

For this problem, you are to write a program that can compute the last non-zero digit of the factorial for N. For example, if your program is asked to compute the last nonzero digit of 5!, your program should produce "2" because 5! = 120, and 2 is the last nonzero digit of 120.

Input

Input to the program is a series of nonnegative integers, each on its own line with no other letters, digits or spaces. For each integer N, you should read the value and compute the last nonzero digit of N!.

Output

For each integer input, the program should print exactly one line of output containing the single last non-zero digit of N!.

Sample Input

Sample Output

  
   1
2
4
8
2
8
  
  
   http://blog.sina.com.cn/s/blog_59e67e2c0100a7yx.html
  
  
   #include<stdio.h>
#include<string.h>
int mod[20]={1,1,2,6,4,2,2,4,2,8,4,4,8,4,6,8,8,6,8,2};
char n[1000];
int a[1000];
int main()
{
 int i,c,t,len;
    while(scanf("%s",n)!=EOF)
 {
  t=1;
  len=strlen(n); 
  for(i=0;i<len;i++)
   a[i]=n[len-1-i]-'0';
  while(len)
  {
   len-=!a[len-1];
   t=t*mod[a[1]%2*10+a[0]]%10;  
   for(c=0,i=len-1;i>=0;i--)      
    c=c*10+a[i],a[i]=c/5,c%=5;
  }
  printf("%d/n",t);
 }
 return 0;
}