Last non-zero Digit in N!(阶乘最后非0位)

时间：2014-05-09 04:13:09 收藏：0 阅读：360

Last non-zero Digit in N!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5861 Accepted Submission(s): 1451

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

9999

Sample Output

Source

South Central USA 1997

 1 #include <iostream>
 2 #include <string.h>
 3 #include <stdio.h>
 4 #include <stdlib.h>
 5 #define MAXN 10000
 6 
 7 int lastdigit(char* buf){
 8     const int mod[20]={1,1,2,6,4,2,2,4,2,8,4,4,8,4,6,8,8,6,8,2};
 9     int len=strlen(buf),a[MAXN],i,c,ret=1;
10     if (len==1)
11         return mod[buf[0]-‘0‘];
12     for (i=0;i<len;i++)
13         a[i]=buf[len-1-i]-‘0‘;
14     for (;len;len-=!a[len-1]){
15         ret=ret*mod[a[1]%2*10+a[0]]%5;
16         for (c=0,i=len-1;i>=0;i--)
17             c=c*10+a[i],a[i]=c/5,c%=5;
18     }
19     return ret+ret%2*5;
20 }
21 int main()
22 {
23     char a[1000]="\0";
24     while(scanf("%s",a)!=EOF)
25     {
26         int ans=lastdigit(a);
27         printf("%d\n",ans);
28     }
29     return 0;
30 }

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