HDU-1757-A Simple Math Problem【矩阵快速幂】

最新推荐文章于 2019-05-14 15:56:22 发布

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最新推荐文章于 2019-05-14 15:56:22 发布

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分类专栏：数学文章标签： HDU1757 矩阵快速幂

本文链接：https://blog.youkuaiyun.com/Li_Hongcheng/article/details/79478383

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本文介绍了一个简单的数学问题，即给定一系列参数和两个正整数k和m，如何计算特定函数f(k)对m取模的结果。文章通过矩阵快速幂的方法解决了该问题，并提供了完整的C++代码实现。

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A Simple Math Problem
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

这里写图片描述

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;

typedef long long int LL;

LL mod;

struct Matrix
{
    LL mat[15][15];
    Matrix()
    {
    memset(mat,0,sizeof(mat));
    }
};

Matrix mul (Matrix a,Matrix b)
{
    Matrix res;
    for(int i=0;i<10;i++)
    {
        for(int j=0;j<10;j++)
        {
            for(int k=0;k<10;k++)
            {
                res.mat[i][j]+=a.mat[i][k]*b.mat[k][j];
                res.mat[i][j]%=mod;
            }
        }
    }
    return res;
}

Matrix pow_matrix(Matrix a,LL n)
{
    Matrix res;
    for(int i=0;i<10;i++)
    {
        res.mat[i][i]=1;
    }
    while(n)
    {
        if(n&1)
            res=mul(res,a);
        a=mul(a,a);
        n>>=1;
    }
    return res;

}



int main()
{
    LL k;
   while(~scanf("%lld%lld",&k,&mod))
   {
       Matrix A,B;
       LL f[11];
       for(int i=0;i<10;i++)
       {
           scanf("%lld",&f[i]);
       }
       if(k<10)
       {
           printf("%lld\n",k%mod);
       }
       else
       {
           for(int i=0;i<10;i++)
           {
               A.mat[i][0]=9-i;
           }
           for(int i=0;i<10;i++)
           {
               B.mat[0][i]=f[i];
           }
           for(int i=1;i<10;i++)
           {
               B.mat[i][i-1]=1;
           }
           B=pow_matrix(B,k-9);
           A=mul(B,A);
           printf("%lld\n",(A.mat[0][0])%mod);
       }

   }

}