hdu------(1757)A Simple Math Problem(简单矩阵快速幂)

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2791    Accepted Submission(s): 1659

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

 代码：

//#define LOCAL
#include<cstdio>
#include<cstring>
#define LL __int64
using namespace std;
const int maxn=10;
LL k,m;
int aa[maxn],mat[maxn][maxn];
int ans[maxn][maxn];

void init()
{
    for(int i=0;i<10;i++)
    {
     for(int j=0;j<10;j++)
     {
       if(i==0)
         mat[i][j]=aa[j];
       else
          if(i==j+1)mat[i][j]=1;
       else mat[i][j]=0;
       if(i==j)
           ans[i][j]=1;
       else ans[i][j]=0;
     }
    }
}

void Matrix(int a[][10],int b[][10])
{

     int c[10][10];
     for(int i=0;i<10;i++)
     {
         for(int j=0;j<10;j++)
        {
            c[i][j]=0;
          for(int k=0;k<10;k++)
           {
            c[i][j]=(c[i][j]+a[i][k]*b[k][j])%m;
          }
        }
     }
    for(int i=0;i<10;i++)
    {
      for(int j=0;j<10;j++)
      {
        a[i][j]=c[i][j];
      }
    }
}

void pow(LL n)
{
    while(n>0)
    {
        if(n&1) Matrix(ans,mat);
        n>>=1L;
        if(n==0)break;
        Matrix(mat,mat);
    }
}

int main()
{
    #ifdef LOCAL
     freopen("test.in","r",stdin);
    #endif

  while(scanf("%I64d%I64d",&k,&m)!=EOF)
  {
       for(int i=0;i<10;i++)
       scanf("%d",aa+i);
     if(k<10) printf("%I64d
",k%m);
     else
     {
       init();
       pow(k-9);
       int res=0;
       for(int i=0;i<10;i++)
        res=(res+(10-i-1)*ans[0][i])%m;
       printf("%d
",res);
    }
  }
  return 0;
}