hdu 5667 BestCoder Round #80 矩阵快速幂

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Sequence

Accepts: 59

Submissions: 650

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

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

$\ \ \ \$ 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.

$f_n=\left{\begin{matrix} 1 ,&n=1 \ a^b,&n=2 \ a^bf_{n-1}^cf_{n-2},&otherwise \end{matrix}\right.$

$\ \ \ \$ He gives you 5 numbers n,a,b,c,p,and he will eat $f_n$ foods.But there are only p foods,so you should tell him $f_n$ mod p.

/*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  0F[n-1]  F[n-2] 1    *    1  0  0                         b  0  1hhh-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 = 100040;struct Matrix{    ll ma[3][3];    Matrix()    {        memset(ma,0,sizeof(ma));    }};Matrix mult(Matrix ta,Matrix tb, ll mod){    Matrix tc;    for(int i = 0; i < 3; i++)    {        for(int j = 0; j < 3; j++)        {            tc.ma[i][j] = 0;            for(int k = 0; k < 3; 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 = 0; i < 3; i++)        t.ma[i][i] = 1;    while(n)    {        if(n & 1) t = mult(t,ta,mod);        ta = mult(ta,ta,mod);        n >>= 1;    }    return t;}ll pow_mod(ll a,ll n,ll mod){    ll t = 1;    a %= mod ;    while(n)    {        if(n & 1) t = t*a%mod;        a = a*a%mod;        n >>= 1;    }    return t;}Matrix mat;Matrix an;ll a,b,c;void ini(ll mod){    mat.ma[0][0] = c,mat.ma[0][1] = 1,mat.ma[0][2] = 0;    mat.ma[1][0] = 1,mat.ma[1][1] = 0,mat.ma[1][2] = 0;    mat.ma[2][0] = b,mat.ma[2][1] = 0,mat.ma[2][2] = 1;    an.ma[0][0] = (b+b*c%mod)%mod,an.ma[0][1] = b,an.ma[0][2] = 1;    an.ma[1][0] = b,an.ma[1][1] = 0,an.ma[2][1] = 1;}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-1);        if(n == 1)        {            printf("1\n");        }        else if(n == 2)            printf("%I64d\n",pow_mod(a,b,mod));        else        {            mat = Mat_pow(mat,n-3,mod-1);            mat = mult(an,mat,mod-1);            ll ci = mat.ma[0][0];            //cout << ci <<endl;            printf("%I64d\n",pow_mod(a,ci,mod));        }    }    return 0;}

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