hdu 5667 sequence

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

 

由遞推式可知，fn是a的次冪，設gn=log a fn，則式子可寫成gn=b+gn-1*c+gn-2;可用矩陣乘法

//
//|G(n)  |  |c 1 b|  |G(n-1)|
//|G(n-1)|= |1 0 0|* |G(n-2)|
//|1     |  |0 0 1|  |1     |
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
ll p;
struct node
{
	ll a[3][3];
};
node mul(node &a,node &b)
{
	node t;
	memset(t.a,0,sizeof(t.a));
	for(int i=0;i<3;i++)
	{
		for(int j=0;j<3;j++)
		{
			for(int k=0;k<3;k++)
			{
				t.a[i][j]+=(a.a[i][k]*b.a[k][j])%(p-1);
				t.a[i][j]%=(p-1); 
			}
		}
	}
	return t;
}
node pow(node a,ll n)
{
	node t;
	memset(t.a,0,sizeof(t.a));
	for(int i=0;i<3;i++)
	t.a[i][i]=1;
	while(n)
	{
		if(n&1)
		t=mul(t,a);
		a=mul(a,a);
		n=n/2;
	}
	return t;
}
ll fun(ll a,ll b,ll c)
{
	ll r=a%c,k=1;
	while(b)
	{
		if(b&1)
		k=(k*r)%c;
		r=(r*r)%c;
		b=b/2;
	}
	return k;
}
int main()
{
	ll t,n,a,b,c;
    cin>>t;
	while(t--)
	{
		cin>>n>>a>>b>>c>>p;
		if(n==1)
		{
			cout<<1<<endl;continue;
		}
		if(n==2)
		{
			cout<<fun(a,b,p)<<endl;continue;
		}
		node t;
		t.a[0][0]=c,t.a[0][1]=1,t.a[0][2]=b;
		t.a[1][0]=1,t.a[1][1]=0,t.a[1][2]=0;
		t.a[2][0]=0,t.a[2][1]=0,t.a[2][2]=1;
		node t1;
		t1=pow(t,n-2);
		ll ans=t1.a[0][0]*b+t1.a[0][2];
		ans=fun(a,ans,p);
		cout<<ans<<endl;
	}
}