hdu 1757

A Simple Math Problem

                                                                              Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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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代码：

#include <iostream>

using namespace std;
const int maxn1=10;
int mod;
struct Matrix
{
    long long m[maxn1][maxn1];
};
Matrix P={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,
          0,0,0,0,0,0,0,1,0,0,
          0,0,0,0,0,0,0,0,1,0
          };
Matrix I={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,
          0,0,0,0,0,0,0,0,0,1
          };
Matrix qm(Matrix a,Matrix b)
{
    Matrix c;
    for(int i=0;i<maxn1;i++)
    {
        for(int j=0;j<maxn1;j++)
        {   c.m[i][j]=0;
            for(int k=0;k<maxn1;k++)
            {
                c.m[i][j]+=((a.m[i][k]%mod)*(b.m[k][j]%mod))%mod;
            }
            c.m[i][j]=c.m[i][j]%mod;
        }
    }
    return c;
}
Matrix qp(long long n)
{
    Matrix m=P,b=I;
    while(n>=1)
    {
        if(n&1) b=qm(b,m);
        n=n>>1;
        m=qm(m,m);
    }
    return b;
}
int main()
{
    int a[10];
    int t,sum;
    while(cin>>t>>mod)
    {
        sum=0;
        for(int i=0;i<=9;i++)
        {
            cin>>a[i];
            P.m[0][i]=a[i];
        }
        if(t<9)
        cout<<t%mod<<endl;
        else
        {
            Matrix tmp=qp(t-9);
            for(int j=9;j>=0;j--)
            sum=(sum+tmp.m[0][9-j]*j)%mod;
            cout<<sum<<endl;
        }
    }
    return 0;
}