## Last non-zero Digit in N!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 815 Accepted Submission(s): 344

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

124828

Source

South Central USA 1997

Recommend

JGShining

求阶乘最后一位非零位。因为N很大，那种暴力的思想肯定是行不通的。这道题想了很久，到现在想的也还不是很明白。但是吉大的ACM模板有这个。所以就直接用了。暂时先放一放吧。

``````/*
* e.cpp
*
*  Created on: 2015年2月5日
*/

#include <iostream>
#include <cstdio>
#include <string.h>
#include <cstdlib>

#define MAXN 10000

int lastdigit(char* buf) {
const int mod = { 1, 1, 2, 6, 4, 2, 2, 4, 2, 8, 4, 4, 8, 4, 6, 8, 8, 6,
8, 2 };
int len = strlen(buf), a[MAXN], i, c, ret = 1;
if (len == 1){
return mod[buf - '0'];
}
for (i = 0; i < len; i++){
a[i] = buf[len - 1 - i] - '0';
}
for (; len; len -= !a[len - 1]) {
ret = ret * mod[a % 2 * 10 + a] % 5;
for (c = 0, i = len - 1; i >= 0; i--){
c = c * 10 + a[i];
a[i] = c / 5;
c %= 5;
}
}
return ret + ret % 2 * 5;
}

int main(){
char n[MAXN];
while(scanf("%s",&n)!=EOF){
printf("%d\n",lastdigit(n));
}

return 0;
}``````