UVA 568 - Just the Facts

Description

The expression N!, read as `` N factorial," denotes the product of the firstN positive integers, where N is nonnegative. So, for example,

NN!011122364245120103628800

For this problem, you are to write a program that can compute the last non-zero digit of any factorial for (). 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 not exceeding 10000, each on its own line with no other letters, digits or spaces. For each integerN, 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. Each line of output should contain the valueN, right-justified in columns 1 through 5 with leading blanks, not leading zeroes. Columns 6 - 9 must contain ``-> " (space hyphen greater space). Column 10 must contain the single last non-zero digit ofN!.

Sample Input 

122612531259999

Sample Output 

    1 -> 1    2 -> 2   26 -> 4  125 -> 8 3125 -> 2 9999 -> 8

求n的阶层的最后一位不为0的数

每次不能单单只留最后一位数

因为可能下一位想成加起来能进一

所以要多保留几位

最少要保留五位数

#include <stdio.h>int main(){int n, temp;while (scanf("%d", &n) != EOF) {long long m = 1;for (int i = 1; i <= n; i++) {temp = i;while (temp % 10 == 0)temp /= 10;m *= temp;while (m % 10 == 0)m /= 10;m = m % 100000;}printf("%5d -> %lld\n", n, m % 10);}return 0;}