HDU 1757 矩阵求第n的递推式

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1838    Accepted Submission(s): 1053

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

 

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 using namespace std;

 const int N= ;

 struct node
 {
     __int64 mat[N][N];
 }hxl;
 int A[],k,m;

 void make_first(node *cur)
 {
     int i,j;
     for(i=;i<=;i++)
     for(j=;j<=;j++)
     if(i==j)
     cur->mat[i][j]=;
     else cur->mat[i][j]=;
 }

 struct node cheng(node cur,node now)
 {
     node ww;
     int i,j,k;
     memset(ww.mat,,sizeof(ww.mat));
     for(i=;i<=;i++)
     for(k=;k<=;k++)
     if(cur.mat[i][k])
     {
         for(j=;j<=;j++)
         if(now.mat[k][j])
         {
             ww.mat[i][j]+=cur.mat[i][k]*now.mat[k][j];
             if(ww.mat[i][j]>=m)
             ww.mat[i][j]%=m;
         }
     }
     return ww;
 }

 void power_sum2(int n)
 {
     __int64 sum=;
     int i;
     node cur,now=hxl;
     make_first(&cur);
     while(n)
     {
         if(n&)
         {
             cur=cheng(cur,now);
         }
         n=n>>;
         now=cheng(now,now);
     }
     for(i=;i<=;i++)
     sum= (sum+(-i)*cur.mat[][i])%m;
     printf("%I64d\n",sum);

 }

 void make_ini()
 {
     int i,j;
     if(k<=)
     {
         printf("%d\n",k);
         return;
     }
     memset(hxl.mat,,sizeof(hxl.mat));
     for(i=;i<=;i++)
     hxl.mat[][i]=A[i];

     for(i=;i<=;i++)// 初始化
     for(j=;j<=;j++)
     if(i-j==)
     hxl.mat[i][j]=;

     power_sum2(k-);
 }

 int main()
 {
     int i;
     while(scanf("%d%d",&k,&m)>)
     {
         for(i=;i<=;i++)
         scanf("%d",&A[i]);
         make_ini();
       //  cs();
     }
     return ;
 }

Source

2007省赛集训队练习赛（6）_linle专场

 

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