Law of Commutation HDU

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

       题意：   略。

       思路：  打表+ 不放弃找规律+ 稍微有点数学意识。  

       比赛的时候打了个表，现在看了好像打表的时候就打错了，不过还是接着找了一下规律：

 a 为奇数的时候 只有 1 ，对于偶数情况，打表发现 b —>  n  的情况 比较混乱，似乎没有规律，就 放弃了......

(大概这就是蒻~吧）

正解：   对于偶数情况 （a,b都为偶数)   b<n  的情况比较混乱，直接暴力枚举， b>=n  的时候就可以有公式可循了。

其实仔细想想也是这样，即 考虑 a^b  大概只有 a^b  % 2^n == 0 的时候才有解，要保证 a^b  存在 2^n  即  b>n ,

其实是瞎解释的，可以忽略。

那么b>= n 的时候：~~~~   点击打开链接

#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 longusing 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;}