Law of Commutation HDU - 6189

Law of Commutation  HDU - 6189

            As we all know, operation ''+'' complies with the commutative law. That is, if we arbitrarily select two integers
a and
b
+b always equals to
+a. However, as for exponentiation, such law may be wrong. In this problem, let us consider a modular exponentiation. Give an integer 

n and an integer aa, count the number of integers
b in the range of
,m] which satisfy the equation a


a (mod
m). 

There are no more than
test cases. 

Input
Each test case contains two positive integers
and a seperated by one space in a line.
For all test cases, you can assume that n≤30,1≤a≤1 n≤30,1≤a≤109.

Output

For each test case, output an integer denoting the number of
 

Sample Input

2 3

2 2

Sample Output

1 

2

       題意：   略。

       思路：  打表+ 不放棄找規律+ 稍微有點數學意識。  

       比賽的時候打了個表，現在看了好像打表的時候就打錯了，不過還是接着找了一下規律：

 a 爲奇數的時候 只有 1 ，對於偶數情況，打表發現 b —>  n  的情況 比較混亂，似乎沒有規律，就 放棄了......

(大概這就是蒻~吧）

正解：   對於偶數情況 （a,b都爲偶數)   b<n  的情況比較混亂，直接暴力枚舉， b>=n  的時候就可以有公式可循了。

其實仔細想想也是這樣，即 考慮 a^b  大概只有 a^b  % 2^n == 0 的時候纔有解，要保證 a^b  存在 2^n  即  b>n ,

其實是瞎解釋的，可以忽略。

那麼b>= n 的時候：~~~~   點擊打開鏈接

#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include <bits/stdc++.h>
#include<string>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<algorithm>
#define maxn 10010
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MOD 1000000007
#define ll long long
using namespace std;

ll Pow(ll a,ll b,ll mod)
{
    ll ans=1;
    a=a%mod;
    while(b)
    {
        if(b&1)
            ans=ans*a%mod;
        b>>=1;
        a=a*a%mod;
    }
    return ans%mod;
}
ll Pow1(ll a,ll b)
{
    ll ans=1;
    while(b)
    {
        if(b&1)
            ans*=a;
        b>>=1;
        a=a*a;
    }
    return ans;
}
int main()
{
    ll n,a;
    while(scanf("%lld%lld",&n,&a)!=EOF)
    {
        if(a&1)
        {
            puts("1");
            continue;
        }
        ll mod=Pow1(2,n);
        ll t=Pow1(2,(n-1)/a+1);
        ll z=mod/t-(n-1)/t;
        ll ans=0;
        for(int i=1;i<n;i++)
        {
            if(Pow(a,i,mod)==Pow(i,a,mod))
                ans++;
        }
        ans+=z;
        cout<<ans<<endl;
    }
    return 0;
}