hdu 5667 Sequence【矩阵快速幂】

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Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 285 Accepted Submission(s): 92

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

题目大意：根据给出的公式，求其解fn、

思路：根据公式得出结论，求的的fn，一定是a的多少次方，所以我们锁定思路是求a的幂数。然后再用快速幂求出解。

公式不难推出：

Fn=Fn-1*c+Fn-2+b；

然后我们也不难写出矩阵：

坑点：在矩阵快速幂的时候要注意先对mod-1，再进行。也就是要注意amodp==0的情况。

然后再进行a的这些次方即可。

Ac代码：

#include<stdio.h>#include<iostream>#include<string.h>using namespace std;#define ll long long intll mod;typedef struct Matrix{    ll mat[3][3];} matrix;matrix A,B,tmp;Matrix matrix_mul(matrix a,matrix b){    matrix c;    memset(c.mat,0,sizeof(c.mat));    int i,j,k;    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]+=a.mat[i][k]*b.mat[k][j];                c.mat[i][j]%=mod;            }        }    }    return c;}Matrix matrix_quick_power(matrix a,ll k)//矩阵快速幂0.0{    matrix b;    memset(b.mat,0,sizeof(b.mat));    for(int i=0; i<3; i++)        b.mat[i][i]=1;//单位矩阵b    while(k)    {        if(k%2==1)        {            b=matrix_mul(a,b);            k-=1;        }        else        {            a=matrix_mul(a,a);            k/=2;        }    }    return b;}ll Mod_pow(ll a, ll b, ll p){    a %= p;    ll ans = 1ll;    while (b)    {        if (b & 1)ans = ans*a%p;        a = a*a%p;        b >>= 1;    }    return ans;}int main(){    ll n,a,b,c;    int t;    scanf("%d",&t);    while(t--)    {        scanf("%I64d%I64d%I64d%I64d%I64d", &n, &a, &b, &c, &mod);        A.mat[0][0] = c, A.mat[0][1] = 1, A.mat[0][2] = b;        A.mat[1][0] = 1, A.mat[1][1] = 0, A.mat[1][2] = 0;        A.mat[2][0] = 0, A.mat[2][1] = 0, A.mat[2][2] = 1;        if (n == 1)printf("1\n");        else if (n == 2)printf("%I64d\n", Mod_pow(a, b, mod));        else        {            mod--;            B =  matrix_quick_power(A,(n - 2));            ll tmp=B.mat[0][0]*b+B.mat[0][2];            mod++;            ll ans = Mod_pow(a, tmp, mod);            printf("%I64d\n",ans);        }    }}

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