HDU 5667 Sequence

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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

15 3 3 3 233

Sample Output

190
把次数拿下来就可以用矩阵递推了，要注意会有a%p=0的情况
#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;typedef long long LL;LL n,a,b,c,p;int T;struct Matrix{   LL a[3][3];}A,B;//A,初始矩阵，B，构造矩阵Matrix operator*(const Matrix&a,const Matrix&b)//矩阵乘法加费马小定理{    Matrix 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;}Matrix pow(Matrix a,LL x){    Matrix c;    for(int i=0; i<3; i++)    {        for(int j=0; j<3; j++)        {            if(i==j)c.a[i][j]=1;            else c.a[i][j]=0;        }    }    while(x)    {        if(x&1) c=c*a;        a=a*a;        x>>=1;    }    return c;}//LL get(LL x,LL y)//{//    LL res=x;//    while(y)//    {//        if(y&1)res=(res*x)%p;//        x=((x%p)*(x%p))%p;//        y>>=1;//    }//    return res;//}LL powmod(LL a,LL b,LL c){    if(b==0)b=c-1;//没有这句话，会WA.    LL ans=1;    while (b)    {        if (b%2==1) ans=ans*a%c;        b/=2;        a=a*a%c;    }    return ans;}int main(){    scanf("%d",&T);    while(T--)    {        cin>>n>>a>>b>>c>>p;        A.a[0][0]=0,A.a[0][1]=b,A.a[0][2]=b;        B.a[0][0]=0,B.a[0][1]=1,B.a[0][2]=0;        B.a[1][0]=1,B.a[1][1]=c,B.a[1][2]=0;        B.a[2][0]=0,B.a[2][1]=1,B.a[2][2]=1;        if(n==1){printf("1\n");continue;}        B = pow(B,n-2);        A = A*B;        LL ans = powmod(a,A.a[0][1],p);        cout<<ans<<endl;    }    return 0;}

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