Just the Facts——阶乘计算

来源：互联网发布：linux ping 100个包编辑：程序博客网时间：2024/05/16 18:49

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 ( $0 \le N \le 10000$ ). 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

题意：给出一个数，求这个数的阶乘，输出阶乘结果从低位开始，第一个不为零的数。

一个数无论和多大的数相乘，得出结果的最后几位只需要用该数和大数的最后几位相乘得出。

解析：打表直接输出不就好了？可是long long到了20的阶乘之后就会超范围了，不过存储能够用到的范围内的数不就好了。还是直接上代码吧，感觉挺简单的，也就不能说什么了。

#include <stdio.h>#include <string.h>long long f[10009];void init(){    f[0] = 1;    f[1] = 1;    f[2] = 2;    long long i;    for(i = 3;i <= 10000;i++)    {        long long ji = i * f[i - 1];        long long yu = ji % 10;        ji = ji / 10;        while(!yu)        {            yu = ji % 10;            ji = ji / 10;        }        ji = ji * 10 + yu;        f[i] = ji % 1000000;    }}int main(){    init();    long long i;    while(~scanf("%lld",&i))    {        printf("%5lld -> %lld\n",i,f[i] % 10);    }    return 0;}

0 0