zoukankan      html  css  js  c++  java
  • Codeforces Round #422 (Div. 2) A. I'm bored with life 暴力

    A. I'm bored with life
     
     

    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
    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:

      给你A,B,求出其阶乘的GCD

    题解:

      min(A,B)<=12

      暴力吧

    #include<bits/stdc++.h>
    using namespace std;
    #pragma comment(linker, "/STACK:102400000,102400000")
    #define ls i<<1
    #define rs ls | 1
    #define mid ((ll+rr)>>1)
    #define pii pair<int,int>
    #define MP make_pair
    typedef long long LL;
    const long long INF = 1e18+1LL;
    const double pi = acos(-1.0);
    const int N = 533333+10, M = 1e3+20,inf = 2e9;
    
    
    int a,b;
    int main() {
        scanf("%d%d",&a,&b);
        LL ans = 1;
        for(int i = 1; i <= min(a,b); ++i) {
            ans *= 1LL*i;
        }
        cout<<ans<<endl;
        return 0;
    }
  • 相关阅读:
    泛海精灵软件预发布统计报告 & 反馈
    【scrum】2.23
    VS2010中C#添加图片(资源)
    用XML存储程序的配置
    scrum 2.28
    【Scrum】2.24
    《人月神话》读书心得
    微软R&D喜欢什么人才
    “电脑族”应多做下肢运动
    ASCII表
  • 原文地址:https://www.cnblogs.com/zxhl/p/7133914.html
Copyright © 2011-2022 走看看