UVA 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,

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 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 

12
26
125
3125
9999

Sample Output 

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

---

Miguel A. Revilla
1998-03-10

     题意：每次输入一个数n，求n！的最后一位不为0的数字。

代码：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;

int main()
{
    int a,b;
    int s[10010];
    int q[10010];
    int x,y;
    s[0] = 1;
    s[1] = 1;
    q[0] = 1;
    q[1] = 1;
    for(int i=2;i<10010;i++)
    {
        b = i;
        a = s[i-1];
        a = a * b;
        y = 0;
        int k = 10;
        while(y == 0)
        {
            y = a % k;
            k = k * 10;
        }
        k = k / 100;
        a = a / k;
        a = a % 100000;
        s[i] = a;
    }
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        x = 0;
        int kk = 10;
        int aa  = s[n];
        while(x == 0)
        {
            x = aa % kk;
            kk = kk * 10;
        }
        printf("%5d -> %d\n",n,x);
    }
    return 0;
}