hdu 5667 Sequence(矩阵快速幂+快速幂+费马小定理）（bestcoder# 80 1002)

Problem Description

     Holion August will eat every thing he has found.

     Now there are many foods,but he does not want to eat all of them at once,so he find a sequence.

fn=⎧⎩⎨⎪⎪1,ab,abfcn−1fn−2,n=1n=2otherwise

     He gives you 5 numbers n,a,b,c,p,and he will eat  fn  foods.But there are only p foods,so you should tell him  fn  mod p.

 

Input

     The first line has a number,T,means testcase.

     Each testcase has 5 numbers,including n,a,b,c,p in a line.

    1≤T≤10,1≤n≤1018,1≤a,b,c≤109 , p  is a prime number,and  p≤109+7 .

 

Output

     Output one number for each case,which is  fn  mod p.

 

Sample Input

  

   1
5 3 3 3 233
  

 

Sample Output

  

   190
  

 

Source

BestCoder Round #80

 

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题目大意：

        给定如题目中所述的公式，求出f（n）mod p，p是素数

解题思路：

        观察公式并手写几项下来，可以发现a^b总是一起出现并且只有指数发生了变化，所以设g(n）=log f(n)  (以a^b为底)

        根据公式：g(n)=1+c*g(n-1)+g(n-2) mod phi(p) (n>=2) 注意是phi(p)，那么就可以用矩阵快速幂：(A^表示A的转置）

          （g(n),g(n-1),1)^=(c,1,1;  1,0,0;  0,0,1;)*(g(n-1),g(n-2),1)^ 根据g(1),g(2)的值算出需要的g(n)来。

        之后用快速幂(a^b)^g(n) mod p可以得出f(n)

        注：这种方法没有遇出题人t说的坑点

#include<stdio.h>
#include<stdio.h>
#include<iostream>
#define ll long long
#define rush() int t;scanf("%d",&t);while(t--)
using namespace std; 
struct mat { ll a[3][3]; };
ll mod,mod1;
mat operator * (mat a, mat b)
{
	mat ans;
	memset(ans.a, 0, sizeof(ans.a));
	for (int i = 0; i < 3; i++)
		for (int j = 0; j < 3; j++)
			for (int k = 0; k < 3; k++)
			{
				ans.a[i][j] += (a.a[i][k] * b.a[k][j]) % (mod-1);
				ans.a[i][j] %= mod-1;
			}
	return ans;
}
mat qpow(mat a, ll n)
{
	mat ans;
	memset(ans.a, 0, sizeof(ans.a));
	for (int i = 0; i < 3; i++)ans.a[i][i] = 1;
	while (n)
	{
		if (n & 1)ans = ans*a;
		a = a*a;
		n >>= 1;
	}
	return ans;
}
ll qpow(ll a, ll b)
{
	ll ans = 1;
	while (b)
	{
		if (b & 1)ans = (ans*a)%mod;
		a =(a*a)%mod;
		b >>= 1;
	}
	return ans;
}/*
ll phi(ll n)
{
	ll temp;
	temp = n;
	for (int i = 2; i*i <= n; i++)
	{
		if (n%i == 0)
		{
			while (n%i == 0) n = n / i;
			temp = temp / i*(i - 1);
		}
		if (n<i + 1)
			break;
	}
	if (n>1)
		temp = temp / n*(n - 1);
	return temp;
}*/
int main()
{
	rush()
	{
		mat tem;
		ll a, b, n,c;
		cin >> n >> a >> b >> c>>mod;
		ll k = qpow(a, b);
		if (n == 1) {cout<<'1'<<endl; continue; }
		else if (n == 2) { cout << k << endl; continue; }
		else
		{
			tem.a[0][0] = c;
			tem.a[0][1] = 1;
			tem.a[0][2] = 1;
			tem.a[1][0] = 1;
			tem.a[1][1] = 0;
			tem.a[1][2] = 0;
			tem.a[2][0] = 0;
			tem.a[2][1] = 0;
			tem.a[2][2] = 1;
			tem = qpow(tem, n - 2);//不能用phi(n-2)
			ll ans = ((tem.a[0][0]%(mod-1) + tem.a[0][2]%(mod-1)) % (mod-1) + mod-1) % (mod-1);
			ans = qpow(k, ans);
			cout << ans << endl;
		}
	}
}