HDU 5667 Sequence（矩阵快速幂）

最新推荐文章于 2022-08-12 15:32:06 发布

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最新推荐文章于 2022-08-12 15:32:06 发布

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分类专栏：矩阵快速幂文章标签：矩阵快速幂

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本文介绍了一种利用矩阵快速幂优化斐波那契型数列求解的方法，针对大规模数据输入提供了高效解决方案。通过将递推公式转化为矩阵形式，并运用快速幂运算技巧，有效地解决了指数级增长带来的计算瓶颈。

Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1916 Accepted Submission(s): 638

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

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

Source

BestCoder Round #80

Recommend

wange2014

题意：给出公式，求Fn。

思路：我们可以把 F [ 1 ] 看作是a^b的0次方，把 F [ 2 ] 看作是a^b的一次方，F [ n ] = a^b 的 n - 1 次方，这样一来我们可以列出等式 (a^b)^(n - 1) = (a^b)^( (n - 2) * c + (n - 3) + 1)，我们可以对 a 的指数用矩阵快速幂求解，化简可得 Fn = Fn-1*c + Fn-2 + b ，需要注意的是 a 的指数太大，需要用欧拉降幂优化一下。

#include <bits/stdc++.h>
using namespace std;
const int N = 3 ;
long long  n,a,b,c,p;
struct xx
{
    long long a[N][N];
} ori;
xx mul(xx x,xx y)
{
    xx temp;
    memset(temp.a,0,sizeof(temp.a));
    for(int i=0; i<N; i++)
    {
        for(int j=0; j<N; j++)
        {
            for(int k=0; k<N; k++)
            {
                temp.a[i][j]+=x.a[i][k]*y.a[k][j]%p;
            }
        }
    }
    return temp;
}
void inti()
{
    int f[][N]={c,1,0,1,0,0,b,0,1};
    for(int i=0;i<N;i++)
    {
        for(int j=0;j<N;j++)
        {
            ori.a[i][j]=f[i][j];
        }
    }
}
xx calc(long long k)
{
    xx b;
    memset(b.a,0,sizeof(b.a));
    for(int i=0; i<3; i++)
        b.a[i][i]=1;
    while(k)
    {
        if(k%2==1)
        {
            b=mul(ori,b);
            k-=1;
        }
        else
        {
            ori=mul(ori,ori);
            k/=2;
        }
    }
    return b;
}
long long  Pow(long long a, long long b)
{
    a %= p;
    long long ans = 1ll;
    while (b)
    {
        if (b & 1)ans = ans*a%p;
        a = a*a%p;
        b >>= 1;
    }
    return ans;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld%lld%lld%lld",&n,&a,&b,&c,&p);
        if(n==1) printf("1\n");
        else if(n==2) printf("%lld\n",Pow(a,b));
        else
        {
            inti();
            p--;
            xx C=calc(n-2);
            p++;
            long long tmp=C.a[0][0]*b+C.a[2][0];
            long long ans=Pow(a,tmp)%p;
            printf("%lld\n",ans);
        }
    }
}