HDU 5667 矩阵快速幂关于指数的递推

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Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1498 Accepted Submission(s): 505

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

根据题意写成递推式，然后可以看出

指数有递推式，可以通过矩阵快速幂来求解。再用下面这公式快速幂取模即可。

（C是素数）

C是素数所以 x只需要对(C-1) 取模就行了

对于指数，求可以得到递推式
f1=0;
f2=b;
f3=c*b;
f(n)=c*f(n-1)+f(n-2)+b;
/*
| c 1 b| | Fn-1| | Fn |
| 1 0 0| * | Fn-2| -》 | Fn-1|
| 0 0 1| | 1 | | 1 |
*/

#include<bits/stdc++.h>#define LL long long#define INF 0x3f3f3f3f#define mem(a,b) memset(a,b,sizeof(a))#define bug puts("***********")using namespace std;const int N =510;LL mod,eul;struct Mat{    LL mat[3][3];}sta,fi;Mat mul(Mat a,Mat b){    Mat c;    mem(c.mat,0);    for(int i=0;i<3;i++){        for(int j=0;j<3;j++){            for(int k=0;k<3;k++){                c.mat[i][j]=(c.mat[i][j]+a.mat[i][k]*b.mat[k][j]%(mod-1))%(mod-1);            }        }    }    return c;}Mat quick(Mat a,LL k){    Mat c;    mem(c.mat,0);    for(int i=0;i<3;i++)        c.mat[i][i]=1;    while(k){        if(k&1) c=mul(c,a);        a=mul(a,a);        k>>=1;    }    return c;}LL quick(LL a,LL k){    LL ans=1;    while(k){        if(k&1)ans=(ans*a)%mod;        a=(a*a)%mod;        k>>=1;    }    return ans%mod;}int main(){    int t;//    cout<<(LL)pow(201,3)*27*27%233<<endl;//    cout<<(LL)pow(220,3)*27*201%233<<endl;    scanf("%d",&t);    LL n,a,b,c;    while(t--){        scanf("%lld%lld%lld%lld%lld",&n,&a,&b,&c,&mod);        mem(sta.mat,0);        mem(fi.mat,0);        LL temp=quick(a,b);        fi.mat[2][0]=1;        fi.mat[1][0]=temp;        fi.mat[0][0]=temp*quick(temp,c)%mod;        sta.mat[0][0]=c;        sta.mat[0][1]=1;        sta.mat[0][2]=b;        sta.mat[1][0]=sta.mat[2][2]=1;        if(n<=3){            printf("%lld\n",fi.mat[3-n][0]);            continue;        }      //  puts("***");        sta=quick(sta,n-2);        LL ans=sta.mat[0][0]*b+sta.mat[0][2];          //cout<<fi.mat[0][0]<<endl;        ans=quick(a,ans);        printf("%lld\n",ans);    }    return 0;}

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