UVA 568 Just the Facts

  Just the Facts 

The expression N!, read as ``N factorial," denotes the product of the first N 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 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. Each line of output should contain the value N, 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 of N!.

Sample Input 

122612531259999

Sample Output 

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

输入一个数字。求出这个数字的阶乘的非零末位。

由于数据只有1W。可以一位一位乘过去。。但是如果整个数都保存的话会超过范围。所以每次只保留5位数

#include <stdio.h>#include <string.h>int n;int main(){    while (scanf("%d", &n) != EOF)    {int sb = 1;for (int i = 2; i <= n; i ++){    sb *= i;    while (sb % 10 == 0)    {sb /= 10;    }    sb %= 100000;}printf("%5d -> %d\n", n, sb % 10);    }    return 0;}