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=;
 LL k,m;
 int aa[maxn],mat[maxn][maxn];
 int ans[maxn][maxn];

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

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

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

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

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

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