hdu 1757 A Simple Math Problem【矩阵快速幂】

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3880    Accepted Submission(s): 2342

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

 

盗一个图：

问题直接简单明了0.0

AC代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
int n,mod;
typedef struct Matrix
{
    int mat[10][10];
}matrix;
matrix A,B;
Matrix matrix_mul(matrix a,matrix b)
{
    matrix c;
    memset(c.mat,0,sizeof(c.mat));
    int i,j,k;
    for(int i=0;i<10;i++)
    {
        for(int j=0;j<10;j++)
        {
            for(int k=0;k<10;k++)
            {
                c.mat[i][j]+=a.mat[i][k]*b.mat[k][j];
                c.mat[i][j]%=mod;
            }
        }
    }
    return c;
}
Matrix matrix_quick_power(matrix a,int k)//矩阵快速幂0.0
{
    matrix b;
    memset(b.mat,0,sizeof(b.mat));
    for(int i=0;i<10;i++)
    b.mat[i][i]=1;//单位矩阵b
    while(k)
    {
        if(k%2==1)
        {
            b=matrix_mul(a,b);
            k-=1;
        }
        else
        {
            a=matrix_mul(a,a);
            k/=2;
        }
    }
    return b;
}
int main()
{
    while(cin>>n>>mod)
    {
        for(int i=0;i<10;i++)
        {
            scanf("%d",&A.mat[0][i]);
        }
        for(int i=1;i<10;i++)
        A.mat[i][i-1]=1;
        B=matrix_quick_power(A,n-9);
        int sum=0;
        for(int i=0;i<10;i++)
        {
            sum+=B.mat[0][i]*(10-i-1);
            sum%=mod;
        }
        printf("%d\n",sum);
    }
}