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
1
2
26
125
3125
9999
Sample Output
1
2
4
8
2
8
Source
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 }