CSU - 1594 Factorials

题目：

Description

The factorial of an integer N, written N!, is the product of all the integers from 1 through N inclusive. The factorial quickly becomes very large: 13! is too large to store in a 32-bit integer on most computers, and 70! is too large for most floating-point variables. Your task is to find the rightmost non-zero digit of n!. For example, 5! = 1 * 2 * 3 * 4 * 5 = 120, so the rightmost non-zero digit of 5! is 2. Likewise, 7! = 1 * 2 * 3 * 4 * 5 * 6 * 7 = 5040, so the rightmost non-zero digit of 7! is 4.

Input

This problem includes multiple cases. The first line of input is a integer T represents the number of cases. 
For each case, there is a single positive integer N no larger than 4,220.

Output

For each case, output corresponding results separately.
The output of each case is a single line containing but a single digit: the right most non-zero digit of N!.

Sample Input

1
7

Sample Output

4

有个题目是这个题目的直接推广，我先做了那个题目，所以这个题目秒了。

点击打开链接（SCU - 1115 阶乘）

代码：

#include<iostream>
using namespace std;

int degree_in_fact(int m, int p)
{
	if (m)return degree_in_fact(m / p, p) + m / p;
	return 0;
}

int main()
{
	int k, n;
	cin >> k;
	while (k--)
	{
		cin >> n;
		int num = degree_in_fact(n, 5);
		int fact = 1;
		for (int i = 1; i <= n; i++)
		{
			int j = i;
			while (j % 2 == 0 && num)
			{
				j /= 2;
				num--;
			}
			while (j % 5 == 0)j /= 5;
			fact = fact*j % 10;
		};
		cout << fact << endl;
	}
	return 0;
}