HDU 5667 Sequence（矩阵快速幂）

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

用矩阵快速幂的时候，注意对p-1取余
递推式：a[n]=c*a[n-1]+a[n-2]+1;

HDU 5667 Sequence（矩阵快速幂）

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>

using namespace std;
typedef long long int LL;
struct Node
{
    LL a[3][3];
}A,B,C;
LL p,n,a,b,c;
Node multiply(Node a,Node b)
{
    Node c;
    for(int i=0;i<3;i++)
    {
        for(int j=0;j<3;j++)
        {
            c.a[i][j]=0;
            for(int k=0;k<3;k++)
            {
                (c.a[i][j]+=(a.a[i][k]*b.a[k][j])%(p-1))%=(p-1);
            }
        }
    }
    return c;
}
Node get(Node a,LL x)
{
    Node c;
    for(int i=0;i<3;i++)
        for(int j=0;j<3;j++)
             c.a[i][j]=(i==j?1:0);
    for(x;x;x>>=1)
    {
        if(x&1) c=multiply(c,a);
        a=multiply(a,a);
    }
    return c;
}
LL quick(LL x,LL y)
{
    if(n>1&&y==0) y=p-1;
    LL ans=1;
    for(y;y;y>>=1)
    {
        if(y&1)  ans=(ans*x)%p;
        x=(x*x)%p;
    }
    return ans;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld%lld%lld%lld",&n,&a,&b,&c,&p);
        A.a[0][0]=0;A.a[1][0]=0;A.a[2][0]=1;
        B.a[0][0]=c; B.a[0][1]=1; B.a[0][2]=1;
        B.a[1][0]=1; B.a[1][1]=0; B.a[1][2]=0;
        B.a[2][0]=0; B.a[2][1]=0; B.a[2][2]=1;
        if(n==1) {cout<<1<<endl;continue;}
        B=get(B,n-1);
        B=multiply(B,A);
        LL num=((B.a[0][0]%(p-1))*(b%(p-1)))%(p-1);
        //cout<<num<<endl;
        cout<<quick(a,num)<<endl;
    }
    return 0;
}