HDU1757 A Simple Math Problem——————矩阵快速幂

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本文介绍了一种使用矩阵快速幂技术解决大规模递归数列计算问题的方法，特别适用于当递归深度非常大，直接递归计算效率极低的情况。通过构造特定的矩阵并运用快速幂运算，可以高效地求解特定形式的递归数列，如题目中给出的复杂递推公式。

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6573 Accepted Submission(s): 4033

Problem Description

Lele now is thinking about a simple function f(x).

If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
And ai(0<=i<=9) can only be 0 or 1 .

Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m.

Input

The problem contains mutiple test cases.Please process to the end of file.
In each case, there will be two lines.
In the first line , there are two positive integers k and m. ( k<2*10^9 , m < 10^5 )
In the second line , there are ten integers represent a0 ~ a9.

Output

For each case, output f(k) % m in one line.

Sample Input

  
   10 9999
1 1 1 1 1 1 1 1 1 1
20 500
1 0 1 0 1 0 1 0 1 0

Sample Output

  
   45
104

Author

linle

Source

2007省赛集训队练习赛（6）_linle专场

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矩阵快速幂解决
第一步，构造矩阵：
$\begin{pmatrix} a_0 & a_1 & a_2 & a_3 & a_4& a_5 & a_6& a_7 & a_8&a_9\\ 1&0 &0 &0 &0 &0 &0 &0 &0 &0 \\ 0 &1 &0 &0 &0 &0 &0 &0 &0 &0 \\ 0&0 &1 &0 &0 &0 &0 &0 &0 &0 \\ 0&0 &0 &1 &0 &0 &0 &0 &0 &0 \\ 0&0 &0 &0 &1 &0 &0 &0 &0 &0 \\ 0&0 &0 &0 &0 &1 &0 &0 &0 &0 \\ 0&0 &0 &0 &0 &0 &1 &0 &0 &0 \\ 0&0 &0 &0 &0 &0 &0 &1 &0 &0 \\ 0&0 &0 &0 &0 &0 &0 &0 &1 &0 \\ \end{pmatrix}\times \begin{pmatrix}f_{x-1}\\f_{x-2}\\f_{x-3}\\f_{x-4}\\f_{x-5}\\f_{x-6}\\f_{x-7}\\f_{x-8}\\f_{x-9}\\f_{x-10}\end{pmatrix} =\begin{pmatrix}f_{x}\\ f_{x-1}\\f_{x-2}\\f_{x-3}\\f_{x-4}\\f_{x-5}\\f_{x-6}\\f_{x-7}\\f_{x-8}\\f_{x-9}\end{pmatrix}$
然后利用矩阵快速幂求得：
$\begin{pmatrix} a_0 & a_1 & a_2 & a_3 & a_4& a_5 & a_6& a_7 & a_8&a_9\\ 1&0 &0 &0 &0 &0 &0 &0 &0 &0 \\ 0 &1 &0 &0 &0 &0 &0 &0 &0 &0 \\ 0&0 &1 &0 &0 &0 &0 &0 &0 &0 \\ 0&0 &0 &1 &0 &0 &0 &0 &0 &0 \\ 0&0 &0 &0 &1 &0 &0 &0 &0 &0 \\ 0&0 &0 &0 &0 &1 &0 &0 &0 &0 \\ 0&0 &0 &0 &0 &0 &1 &0 &0 &0 \\ 0&0 &0 &0 &0 &0 &0 &1 &0 &0 \\ 0&0 &0 &0 &0 &0 &0 &0 &1 &0 \\ \end{pmatrix}^{n-9}\times \begin{pmatrix}f_{9}\\f_{8}\\f_{7}\\f_{6}\\f_{5}\\f_{4}\\f_{3}\\f_{2}\\f_{1}\\f_{0}\end{pmatrix} =\begin{pmatrix}f_{n}\\ f_{n-1}\\f_{n-2}\\f_{n-3}\\f_{n-4}\\f_{n-5}\\f_{n-6}\\f_{n-7}\\f_{n-8}\\f_{n-9}\end{pmatrix}$

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN =  13;

struct mat{ll m[MAXN][MAXN]; };

ll n,MOD;
ll N,k,b_n=0;
ll C[MAXN],h[MAXN];

mat mul(mat a,mat b)//矩阵乘法
{
        mat tmp;
        for(int i=1;i<=N;i++)
                for(int j=1;j<=N;j++)
                {
                        tmp.m[i][j]=0;
                        for(int k=1;k<=N;k++)
                                tmp.m[i][j]=(tmp.m[i][j]+a.m[i][k]*b.m[k][j]+MOD)%MOD;
                }
        return tmp;
}
mat pow_mod(mat a,ll n)
{
        mat ans;
        for(int i=1;i<=N;i++)//构造单位矩阵
                for(int j=1;j<=N;j++)
                        ans.m[i][j]=(i==j);
        while(n)//矩阵快速幂
        {
                if(n&1) ans=mul(ans,a);
                a=mul(a,a);
                n>>=1;
        }
        return ans;
}

void init(mat &res,mat &H)//构造矩阵
{
        for(int i=1;i<=N;i++)
                res.m[1][i]=C[i];
        res.m[1][N]=b_n;
        for(int i=2;i<=N;i++)
                for(int j=1;j<=N;j++)
                        res.m[i][j]=(i==j+1);
        res.m[N][N-1] = 0;
        res.m[N][N]   = 1;
        for(int i=1;i<=N;i++)
        {
                H.m[i][1]=h[N-i];
                for(int j=2;j<=N;j++)
                        H.m[i][j]=0;
        }
        H.m[N][1]=1;
}
void slove(ll k,ll n)
{
        mat res,H;
        init(res,H);
        res = pow_mod(res,n-k+1);
        res = mul(res,H);
        ll ans=res.m[1][1];

        printf("%lld\n",ans%MOD);
}

int main()
{
        k=10; N=11;
        for(int i=1;i<=k;i++)   h[i]=i-1;

        while(~scanf("%lld %lld",&n,&MOD))//f(n)%MOD;,我更习惯这样写
        {

                for(int i=1;i<=k;i++)
                        scanf("%lld",&C[i]);//第一次没有加&，找了好久。。。我好傻
                if(n<10)
                {
                        printf("%d\n",n);
                        continue;
                }
                slove(k,n);
        }
        return 0;
}