hdu 5667 BestCoder Round #80 矩阵快速幂

Sequence

 Accepts: 59

 Submissions: 650

 Time Limit: 2000/1000 MS (Java/Others)

 Memory Limit: 65536/65536 K (Java/Others)

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.

f_n=\left{\begin{matrix} 1 ,&n=1 \ a^b,&n=2 \ a^bf_{n-1}^cf_{n-2},&otherwise \end{matrix}\right.f​n​​=​⎩​⎨​⎧​​​1,​a​b​​,​a​b​​f​n−1​c​​f​n−2​​,​​​n=1​n=2​otherwise​​

\ \ \ \    He gives you 5 numbers n,a,b,c,p,and he will eat f_nf​n​​ foods.But there are only p foods,so you should tell him f_nf​n​​ 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\le T \le 10,1\le n\le 10^{18},1\le a,b,c\le 10^9    1≤T≤10,1≤n≤10​18​​,1≤a,b,c≤10​9​​,pp is a prime number,and p\le 10^9+7p≤10​9​​+7.

Output

\ \ \ \    Output one number for each case,which is f_nf​n​​ mod p.

Sample Input

1
5 3 3 3 233

Sample Output

190

/*
hdu 5667 BestCoder Round #80 矩阵快速幂
 
F[n] = 1 (n == 1)
F[n] = a^b (n == 2)
F[n] = a^b * F[n-1]^c *F [n-2]
 
最开始试了下化简公式,但是无果. 也从矩阵快速幂上面考虑过(毕竟 F[n]与 F[n-1],F[n-2]有关)
但是发现是 乘法运算不知道怎么弄了(2b了)
 
能够发现运算时基于a的次方的,当a的次方相乘时就变成了他们的次方相加 (好气 TAT)
于是乎 a^g[n] = a^(b + c*g[n-1] * g[n-2])
然后用类似快速幂求斐波那契数的方法即可
 
F[n]    F[n-1] 1         C  1  0
F[n-1]  F[n-2] 1    *    1  0  0
                         b  0  1
hhh-2016-04-18 20:36:40
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#include <algorithm>
#include <functional>
#include <math.h>
using namespace std;
#define lson  (i<<1)
#define rson  ((i<<1)|1)
typedef long long ll;
const int maxn = ;
 
struct Matrix
{
    ll ma[][];
    Matrix()
    {
        memset(ma,,sizeof(ma));
    }
};
 
Matrix mult(Matrix ta,Matrix tb, ll mod)
{
    Matrix tc;
    for(int i = ; i < ; i++)
    {
        for(int j = ; j < ; j++)
        {
            tc.ma[i][j] = ;
            for(int k = ; k < ; k++){
                tc.ma[i][j] += (ta.ma[i][k] * tb.ma[k][j])%mod;
                tc.ma[i][j] %= mod;
            }
        }
    }
    return tc;
}
 
Matrix Mat_pow(Matrix ta,ll n,ll mod)
{
    Matrix t;
    for(int i = ; i < ; i++)
        t.ma[i][i] = ;
    while(n)
    {
        if(n & ) t = mult(t,ta,mod);
        ta = mult(ta,ta,mod);
        n >>= ;
    }
    return t;
}
 
ll pow_mod(ll a,ll n,ll mod)
{
    ll t = ;
    a %= mod ;
    while(n)
    {
        if(n & ) t = t*a%mod;
        a = a*a%mod;
        n >>= ;
    }
    return t;
}
 
Matrix mat;
Matrix an;
ll a,b,c;
void ini(ll mod)
{
    mat.ma[][] = c,mat.ma[][] = ,mat.ma[][] = ;
    mat.ma[][] = ,mat.ma[][] = ,mat.ma[][] = ;
    mat.ma[][] = b,mat.ma[][] = ,mat.ma[][] = ;
 
    an.ma[][] = (b+b*c%mod)%mod,an.ma[][] = b,an.ma[][] = ;
    an.ma[][] = b,an.ma[][] = ,an.ma[][] = ;
}
ll mod,n;
 
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%I64d%I64d%I64d%I64d%I64d",&n,&a,&b,&c,&mod);
        a%=mod,b%=mod,c%=mod;
        ini(mod-);
        if(n == )
        {
            printf("1\n");
        }
        else if(n == )
            printf("%I64d\n",pow_mod(a,b,mod));
        else
        {
            mat = Mat_pow(mat,n-,mod-);
            mat = mult(an,mat,mod-);
            ll ci = mat.ma[][];
            //cout << ci <<endl;
            printf("%I64d\n",pow_mod(a,ci,mod));
        }
    }
    return ;
}