hdu 5667 Sequence【矩阵快速幂】

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 int
ll 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);
        }
    }
}