Law of Commutation HDU - 6189(推导)

最新推荐文章于 2019-04-10 12:46:15 发布

原创最新推荐文章于 2019-04-10 12:46:15 发布 · 325 阅读

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本文探讨了特定条件下模指数运算的性质及其求解方法，重点分析了当底数为偶数时，如何通过数学推导找到满足等式的整数解的数量。

As we all know, operation ”+” complies with the commutative law. That is, if we arbitrarily select two integers a and b, a+b always equals to b+a. However, as for exponentiation, such law may be wrong. In this problem, let us consider a modular exponentiation. Give an integer m=2n and an integer a, count the number of integers b in the range of [1,m] which satisfy the equation $a^b≡b^a (mod\ m).$
Input
There are no more than 2500 test cases.

Each test case contains two positive integers n and a seperated by one space in a line.

For all test cases, you can assume that $n≤30,1≤a≤10^9.$
Output
For each test case, output an integer denoting the number of b.
Sample Input
2 3
2 2
Sample Output
1
2
回归日，哒哒~
观察发现当a为奇数时答案均为1。
当a为偶数时：
则记：

a = 2 y

$a=2y$
则：

a b = 2 b y b

$a^b=2^by^b$
因为n<30,所以当

b<nb<n $b<n$ 时，计算机暴力验证
当

b>=nb>=n $b>=n$ ,易知

a b \equiv 0 (m o d 2 n)

$a^b\equiv0(mod\ 2^n)$
现在看

baba $b^a$ ,易知

b = 2 x y

$b=2^xy$
则

b a = 2 x a y a

$b^a=2^{xa}y^a$
当

xa>=nxa>=n $xa>=n$ 均满足。

#include<iostream>
#include<cstdio>
#include<set>
using namespace std;
long long P(long long a,long long b,long long mod)
{
    long long ans=1;
    while(b)
    {
        if(b&1)ans=ans*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return ans;
}
long long p(long long a,long long b)
{
    long long ans=1;
    while(b)
    {
        if(b&1)ans*=a;
        a*=a;
        b>>=1;
    }
    return ans;
}
int main()
{
    long long a,n;
    while(scanf("%lld%lld",&n,&a)==2)
    {
        if(a&1)cout<<1<<endl;
        else
        {
            long long m=1;
            for(int i=0;i<n;i++)
                m*=2;
            long long ans=0;
            for(int i=1;i<n;i++)
                if(P(a,i,m)==P(i,a,m))ans++;
            //cout<<"ans1: "<<ans<<endl;
            long long now=1;
            for(int i=1;i<=n;i++)
            {
                now*=2;
                if(i*a>=n)
                {
                    //cout<<"i*a: "<<i*a<<endl;
                    long long temp=1;
                    for(;;)
                    {
                        if(temp*now<n)ans--,temp+=2;
                        else break;
                    }//可能包含了前面暴力的部分，所以需要减掉
                    long long res=p(2,n-i);
                    if(res>1)ans+=res/2;
                    else ans++;
                }
            }
            cout<<ans<<endl;
        }
    }
    return 0;
}