1199: Just the Facts 求N!的非零末位

本文介绍了一个程序设计问题,即如何计算任意非负整数阶乘后的尾部第一个非零数字。输入为一系列不超过10000的非负整数,输出则是每个整数阶乘结果的最后一个非零数字。

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1199: Just the Facts


StatusIn/OutTIME LimitMEMORY LimitSubmit TimesSolved UsersJUDGE TYPE
stdin/stdout3s8192K734332Standard

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

 

NN!
01
11
22
36
424
5120
103628800

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

1
2
26
125
3125
9999

 

Sample Output

    1 -> 1
    2 -> 2
   26 -> 4
  125 -> 8
 3125 -> 2
 9999 -> 8
#include<stdio.h>
int main()
{
    int i;
    int n;
    while(scanf("%d",&n)==1)
    {
        printf("%5d -> ",n);
        if(n==0||n==1) {printf("1/n");continue;}
        int temp=1;
        for(i=2;i<=n;i++)
        {
            int b=i;
            while(b%10==0) b/=10;
            while(b%5==0)
            {
                b/=5;
                temp/=2;
            }
            temp*=b;
            temp%=100000;
        }
        printf("%d/n",temp%10);
    }
    return 0;
}
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