codeforces 822A I'm bored with life

最新推荐文章于 2019-07-19 23:48:00 发布

原创最新推荐文章于 2019-07-19 23:48:00 发布 · 425 阅读

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本文介绍了一个简单的算法问题，即求两个数阶乘的最大公因数，并给出了一种有效的解决方案。通过寻找较小数值的阶乘来快速得出结果。

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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 xis 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
        
4 3

          output
        
6

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a！和b！的最大公共因数，我刚开始没想明白，后来才明白的，举个例子，5和6吧，6！=5！*6，这样就明白了吧，大的阶乘一定是小的阶乘再乘其它不为0的数，这样就转化为求小的阶乘。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
using namespace std;
typedef long long ll;
int main()
{
    int a,b;
    cin>>a>>b;
    a=min(a,b);
    int ans=1;
    for(int i=2;i<=a;i++)
        ans*=i;
    cout<<ans<<endl;
    return 0;
}