zoukankan      html  css  js  c++  java
  • Exponial

    Description

    Everybody loves big numbers (if you do not, you might want to stop reading at this point). There are many ways of constructing really big numbers known to humankind, for instance:

    • Exponentiation: 422016=4242...42��2016 times422016=42⋅42⋅...⋅42⏟2016 times.
    • Factorials: 2016!=2016 ⋅ 2015 ⋅ ... ⋅ 2 ⋅ 1.

    Illustration of exponial(3) (not to scale), Picture by C.M. de Talleyrand-Périgord via Wikimedia Commons

    In this problem we look at their lesser-known love-child the exponial, which is an operation defined for all positive integers n as 
    exponial(n)=n(n − 1)(n − 2)21
    For example, exponial(1)=1 and exponial(5)=54321 ≈ 6.206 ⋅ 10183230 which is already pretty big. Note that exponentiation is right-associative: abc = a(bc).

    Since the exponials are really big, they can be a bit unwieldy to work with. Therefore we would like you to write a program which computesexponial(n) mod m (the remainder of exponial(n) when dividing by m).

    Input

    There will be several test cases. For the each case, the input consists of two integers n (1 ≤ n ≤ 109) and m (1 ≤ m ≤ 109).

    Output

    Output a single integer, the value of exponial(n) mod m.

    Sample Input

    2 42
    5 123456789
    94 265

    Sample Output

    2
    16317634
    39

    题解:题意很好理解;直接说题。这是利用欧拉函数降幂公式求的,标准模版(记得开 long long);看代码()

    AC代码为:(我下面有一篇讲下欧拉函数降幂公式)


    #include<iostream>
    #include<algorithm>
    #include<cstring>
    using namespace std;


    typedef long long LL;
    LL N,M;


    LL eular(LL m)
    {
    LL res=m,a=m;
    for(LL i=2;i*i<=a;i++)
    {
    if(a%i==0)
    {
    res=res/i*(i-1);
    while(a%i==0)
    a/=i;
    }
    }
    if(a>1) res=res/a*(a-1);
    return res;
    }


    LL Fast_mod(LL x,LL n,LL m)
    {
    LL res=1;
    while(n>0)
    {
    if(n & 1) res=(res*x)%m;
    x=(x*x)%m;
    n/=2;
    }
    return res;
    }


    LL work(LL n,LL m)
    {
    LL ans;
    if(m==1) return 0;
    else if(n==1) return 1;
    else if(n==2) return 2%m;
    else if(n==3) return 9%m;
    else if(n==4) return Fast_mod(4,9,m);
    else
    {
    LL phi=eular(m);
    LL z=work(n-1,phi);
    ans=Fast_mod(n,phi+z,m);
    }
    return ans;
    }


    int main()
    {
    cin>>N>>M;
    cout<<work(N,M)<<endl;
    return 0;
    }



  • 相关阅读:
    [题解]北京2018
    [数据结构][字典树]Word Puzzles
    [数据结构][字典树]Hardwood Species
    [数学][广义欧拉定理]上帝与集合的正确用法
    Equal Sums
    Useful Decomposition
    网络流 EK算法
    线段树各类操作
    唯一分解定理
    Kuro and Walking Route
  • 原文地址:https://www.cnblogs.com/csushl/p/9386580.html
Copyright © 2011-2022 走看看