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

由递推式可知，fn是a的次幂，设gn=log a fn，则式子可写成gn=b+gn-1*c+gn-2;可用矩阵乘法

//
//|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;
}
}