hdu-5667 Sequence(矩阵快速幂+费马小定理+快速幂)

题目链接：

Sequence

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.

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≤10^18,1≤a,b,c≤10^9,p is a prime number,and p≤10^9+7.

 

Output

 

Output one number for each case,which is fn mod p.

 

Sample Input

 

1

5 3 3 3 233

 

Sample Output

 

190

 

题意：

 

问f(n)对p取模的结果;

 

思路：

 

(a^b)^p[n]= a^b * ((a^b)^p[n-1])^c * ((a^b)^p[n-2]);递推式子可以这样写;

合并后变为(a^b)^p[n]=(a^b)^{(c*p[n-1]+p[n-2]+1);}

可以得到p[n]=c*p[n-1]+p[n-2]+1;

这样的递推式可以用矩阵乘法得到第n项;这是矩阵乘法的一个应用,给matrix67大神的博客地址可以学习,点这里

构造矩阵乘法:

 

p[n]　　　　c    1    1　　　　p[n-1]

p[n-1]   =  1   0    0   *  　p[n-2]

1　　　　　  0   0    1　　　　　　1

 

然后这中间还有一个问题,就是取模的问题;a^b*p[n]%mod=a^{b*p[n]%(mod-1)}%mod;

这是根据费马小定理得到的;a^(p-1)Ξ1%p;这里给出地址

 

a^b*p[n]%mod=a^{b*p[n]/(mod-1)*(mod-1)+b*p[n]%(mod-1)}%mod;

 

令x=b*p[n]/(mod-1)则a^b*p[n]%mod=a^{x*(mod-1)+b*p[n]%(mod-1)}%mod=a^{b*p[n]%(mod-1)}%mod;

这就是取模的结果啦;

然后就是快速幂算答案啦;

还有要注意的就是a%p==0的情况;

 

 

AC代码：

 

 

/*5667    0MS    1584K    1835B    G++    2014300227*/
#include <bits/stdc++.h>
using namespace std;
const int N=1e4+;
typedef long long ll;
const int mod=1e9+;
ll n,a,b,c,p;
struct matrix
{
    ll a[][];
};
matrix A;
void Iint(ll x)//矩阵初始化;
{
    A.a[][]=x;
    A.a[][]=;
    A.a[][]=;
    A.a[][]=;
    A.a[][]=A.a[][]=;
    A.a[][]=A.a[][]=;
    A.a[][]=;
}
matrix mul(matrix x,matrix y)//矩阵相乘
{
    matrix ans;
    for(int i=;i<;i++)
    {
        for(int j=;j<;j++)
        {
            ans.a[i][j]=;
            for(int k=;k<;k++)
            {
                ans.a[i][j]+=(x.a[i][k]*y.a[k][j])%(p-);
                ans.a[i][j]%=(p-);
            }
        }
    }
    return ans;
}
ll fast_fun(matrix temp,ll num)//矩阵快速幂;
{
    matrix s,base;
    for(int i=;i<;i++)//s初始化为单位矩阵
    {
        for(int j=;j<;j++)
        {
            s.a[i][j]=;
            base.a[i][j]=temp.a[i][j];
        }
    }
    s.a[][]=s.a[][]=s.a[][]=;
    while(num)
    {
        if(num&)
        {
            s=mul(s,base);
        }
        base=mul(base,base);
        num=(num>>);
    }
    return (s.a[][]+s.a[][])%(p-);
}
ll fastpow(ll fx,ll fy)//快速幂求结果;
{
    ll s=,base=fx;
    while(fy)
    {
        if(fy&)
        {
            s*=base;
            s%=p;
        }
        base*=base;
        base%=p;
        fy=(fy>>);
    }
    return s;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld%lld%lld%lld",&n,&a,&b,&c,&p);
        Iint(c);
        if(n==)printf("1\n");
        else if(n==)printf("%lld\n",fastpow(a,b));
        else
        {
            if(a%p==)printf("0\n");
            else {
            ll gg=fast_fun(A,n-)*b%(p-);
            printf("%lld\n",fastpow(a,gg));
            }
        }
    }
    return ;
}