HDU 5667 Sequence （BestCoder Round #80 1003）——构造矩阵

版权声明：编写不易，转载请注明出处，谢谢。 https://blog.csdn.net/qingshui23/article/details/51189473

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
   
   

 

Source

BestCoder Round #80

 

题目大意： 

官方题解：

个人的解题思路： 
首先对这个题进行分析一下，我们发现这个f(n)的每一项都有一个底数 a ，所以 我们就对 a 进行取对数，令

g(n)=logaf(n)

即得到一个递推式： 

g(n)=logaab+logaf(n−1)c+logaf(n−2)

即

g(n)=c∗g(n−1)+g(n−2)+b

所以可以构造矩阵 A 满足 

{g(n−1),g(n−2),b}∗A=={g(n),g(n−1),b}

又因为

f(n)=ab∗f(n−1)c∗f(n−2)

所以

g(n)=logaf(n)=b+c∗g(n−1)+g(n−2)

所以构造的矩阵 A 为： 

c11100001

所以 我们可就知道了g(n)的值是多少,当 n=3的时候 
因为： 

{g(2),g(1),b}∗A3−2=={g(3),g(3−1),b}

所以： 

{g(2),g(1),b}∗An−2=={g(n),g(n−1),b}

所以我们只要求吃构造矩阵的的 (n-2)次方来就行了，注意的是取模的时候是对 phi(p)取模因为我们要求的结果是 

f(n)=ag(n)

因为 

ab%mod=ab%phi(mod)+phi(mod)%mod

又因为 p 是质数，所以 phi(p) = p-1; 
这里我们要用到的就是 矩阵的乘法 和 矩阵的快速幂 + 普通的快速幂就行了。

My Code：

#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
const int MAXN = 3;
typedef long long LL;
typedef struct
{
    LL mat[MAXN][MAXN];
} Matrix;
LL c;
///求得的矩阵
Matrix p = {c, 1, 0,
            1, 0, 0,
            1, 0, 1,
           };
///单位矩阵
Matrix I = {1, 0, 0,
            0, 1, 0,
            0, 0, 1,
           };
///矩阵乘法
Matrix Mul_Matrix(Matrix a, Matrix b, LL MOD)
{
    Matrix c;
    for(int i=0; i<MAXN; i++)
    {
        for(int j=0; j<MAXN; j++)
        {
            c.mat[i][j] = 0;
            for(int k=0; k<MAXN; k++)
            {
                c.mat[i][j] += (a.mat[i][k] * b.mat[k][j]) % MOD;
                c.mat[i][j] %= MOD;
            }
        }
    }
    return c;
}
///矩阵的快速幂
Matrix quick_Mod_Matrix(LL m, LL MOD)
{
    Matrix ans = I, b = p;
    while(m)
    {
        if(m & 1)
            ans = Mul_Matrix(ans, b, MOD);
        m>>=1;
        b = Mul_Matrix(b, b, MOD);
    }
    return ans;
}
///普通的快速幂
LL quick_Mod(LL a, LL b, LL MOD)
{
    LL ans = 1;
    while(b)
    {
        if(b & 1)
            ans = (ans * a) % MOD;
        b>>=1;
        a = (a * a) % MOD;
    }
    return ans;
}
int main()
{
    int T;
    LL n, a, b, MOD;
    scanf("%d",&T);
    while(T--)
    {
        cin>>n>>a>>b>>c>>MOD;
        p.mat[0][0] = c;
        if(n == 1)
        {
            puts("1");
            continue;
        }
        if(n == 2)
        {
            LL ret = quick_Mod(a, b, MOD);
            cout<<ret<<endl;
            continue;
        }
        Matrix tmp = quick_Mod_Matrix(n-2, MOD-1);///注意取模是MOD-1
        LL ans = ( b * tmp.mat[0][0] ) % ( MOD-1 ) + ( b * tmp.mat[2][0] ) % ( MOD-1 );
        LL ret = quick_Mod(a, ans, MOD);///a^b%MOD
        cout<<ret<<endl;
    }
    return 0;
}

 

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
    
    

 

Source

BestCoder Round #80