Just the Facts

最新推荐文章于 2019-02-21 10:49:51 发布

原创最新推荐文章于 2019-02-21 10:49:51 发布 · 585 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#高精度

解题报告同时被 2 个专栏收录

71 篇文章

订阅专栏

===高精度===

8 篇文章

订阅专栏

本博客介绍了一个算法，用于计算给定范围内（0≤N≤10000）的阶乘的最后非零数字。通过去除末尾的零并仅关注最后四位数字的乘法，实现高效计算。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

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

For this problem, you are to write a program that can compute the last non-zero digit of any factorial for (0 <= N <= 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

Sample Output

解题思路：

出题人把它归为高精度，真是太坏了！千万别看到阶层就以为是高精度，给的数很吓人，让你求阶层后最右边不为0的数，不会真有人用高精度求了吧。。。。跟高精度八竿子打不着。后面的0乘以任何数都为0，所以不会影响我最后一个不为0的数，唯一担心的是进位问题，由于不会超过10000的阶层，也就是说每次乘的数不会超过10000，我只要每次保证最后一个数不为0，并且取最后4位去乘下一个数就行，无论它怎么进位都影响不到我输出正确答案了。

AC代码：

#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
    int n, ans;
    while(scanf("%d", &n) != EOF)
    {
        ans = 1;
        for(int i = 1; i <= n; i++)
        {
            ans *= i;
            while(ans % 10 == 0)  // 去除末尾的0
                ans /= 10;
            ans %= 10000;  //取最后4位
        }
        printf("%5d -> %d\n", n, ans % 10);  //注意一下输出格式，我PE了一发
    }
    return 0;
}