UVA568 Just the Facts

 Just the Facts 

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

NN!011122364245120103628800

For this problem, you are to write a program that can compute the lastnon-zero digit of anyfactorial for (). For example, if your program is asked tocompute the last nonzerodigit 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 linewith no other letters, digits or spaces. For each integerN, you shouldread the value and compute the last nonzero digit of N!.

Output 

For each integer input, the program should print exactly one line ofoutput. Each line of outputshould contain the valueN, right-justified in columns 1 through 5 withleading blanks, not leadingzeroes. 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题目的意思就是给你一个ｎ　算出ｎ的阶乘的最后一位非０数是多少。。思路就是每乘完一个，就判断求余１０是不是０；如果是就一直除以１０到求余１０不等０为止 。就把０全去掉了。。到最后，求余１０就是最后一位非０数。有一点需要注意，如果只是去掉０，那乘到后面数还是会非常大，会爆掉，所以就一直对１０００００取余，虽然最大到９９９９，但如果是对１００００取余会影响精度。ＡＣ代码：
#include<stdio.h>int main () {int num;int temp;while (scanf("%d",&num) != EOF) {temp = 1;for (int i = 1;i <=num ;i++) {temp = temp * i;while(temp % 10 == 0)temp = temp / 10;temp = temp % 100000;}temp = temp % 10;printf("%5d -> %d\n",num,temp);}return 0;}