HDU-6189 Law of Commutation(数学)

Law of Commutation

                                                                Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                    Total Submission(s): 469    Accepted Submission(s): 172

Problem Description

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  ab≡ba (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≤109.

 

Output

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

 

Sample Input

2 3 2 2

 

Sample Output

1 2

 

题意：输入n,a,令m = 2^n,问b取[1,m],问有多少个这样的b使(a^b)%m=(b^a)%m.

思路：找规律发现当a为奇数的时候，答案恒为1，当a为偶数，a^b为偶数，为使b^a为偶数，b也必须为偶数.

因为n比较小,所以当b小于等于n的时候可以暴力判断，当b> n,(a^b)%m恒为0，所以我们想使(b^a)%m也为0.

因为m = 2^n,所以b^a应为m的倍数,所以b应为2^(n/a)的倍数,此处上取整.所以我们只需算一下m是2^(n/a)的多少倍即可.当然还得减去<= n范围内重复的数量.

代码：

#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
const int maxn = 2e5+5;
const double eps = 1e-12;
const int inf = 0x3f3f3f3f;
map<int,int>::iterator it;

ll n;
ll a,b,mod;

ll quick_pow(ll x,ll y)
{
	ll ans = 1;
	while(y)
	{
		if(y&1)
			ans = ans*x%mod;
		y>>= 1;
		x = x*x%mod;
	}
	return ans;
}

int main()
{
	while(~scanf("%lld %lld",&n,&a))
	{
		if(a&1)
		{
			printf("1\n");
			continue;
		}
		
		mod = 1<<n;
		
		ll ans = 0;
		for(ll i = 2;i<= n;i+= 2)
		{
			if(quick_pow(a,i) == quick_pow(i,a))
				ans++;
		}
		
		ll tmp = (n+a-1)/a;//上取整
		tmp = 1<<tmp;
		ans+= (mod/tmp-n/tmp);//减去重复的
		
		printf("%lld\n",ans);
	}
	
	return 0;
}