Codeforces Round #422 (Div. 2) A. I'm bored with life-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_43537574/article/details/104861460

本文介绍了一种求两个整数阶乘最大公约数的有效算法。通过比较两数大小，利用较小数的阶乘作为答案，实现快速计算。示例代码使用C++实现，输入两整数A和B，输出A!和B!的最大公约数。

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Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!

Leha came up with a task for himself to relax a little. He chooses two integers A and B and then calculates the greatest common divisor of integers "A factorial" and "B factorial". Formally the hacker wants to find out GCD(A!, B!). It's well known that the factorial of an integer x is a product of all positive integers less than or equal to x. Thus x! = 1·2·3·...·(x - 1)·x. For example 4! = 1·2·3·4 = 24. Recall that GCD(x, y) is the largest positive integer q that divides (without a remainder) both x and y.

Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?

Input

The first and single line contains two integers A and B (1 ≤ A, B ≤ 109, min(A, B) ≤ 12).

Output

Print a single integer denoting the greatest common divisor of integers A! and B!.

Example

input

Copy

4 3

output

Copy

Note

Consider the sample.

4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.

题意：求整数A和B阶乘的最大公约数，通过观察发现A和B中较小数的阶乘即为答案

#include<iostream>
using namespace std;
typedef long long ll;
int main()
{
	ll a,b;
	cin>>a>>b; 
	ll ans=1;
	ll x=a;
	if(x>b)
	x=b;
	for(int i=1;i<=x;i++)
	ans*=i;
	cout<<ans<<endl;
	return 0;
	
}