hdu 1757 A Simple Math Problem (乘法矩阵)

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2441    Accepted Submission(s): 1415

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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开始没什么思路，后来还是没思路..

然后看解题报告，发现时乘法矩阵，关键点还是在构造矩阵上。

参考：http://blog.sina.com.cn/s/blog_79b832820100wnu3.html

f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10)

构造的矩阵是:

|0 1 0 ........... 0|    |f0|    |f1 |
|0 0 1 0 ........ 0|    |f1|    |f2 |
|.....................1| * |...| = |...|
|a9 a8 .........a0 |    |f9|    |f10|

设为矩阵 A * 矩阵B =矩阵C

我们要求的是 f(k),就是矩阵C的最后一个元素，故依据矩阵的结合律，可看到

C=A*(A*(.....*(A*B))) ,要有k个A即 C=A^k*B ，然后就可以二分求A^k，最后乘上B就可以求得矩阵C

 //0MS    232K    1214 B    C++
 #include<stdio.h>
 #include<string.h>
 #define N 15
 struct matrix{
     int g[N][N];
 }ans,temp;
 int g[N];
 int m;
 matrix mul(matrix a,matrix b)
 {
     matrix c;
     for(int i=;i<;i++)
         for(int j=;j<;j++){
             c.g[i][j]=;
             for(int k=;k<;k++)
                 c.g[i][j]+=a.g[i][k]*b.g[k][j];
                 c.g[i][j]%=m;
         }
     return c;
 }
 void solve(int k)
 {
     while(k){
         if(k&) ans=mul(temp,ans);
         temp=mul(temp,temp);
         k/=;
     }
     int sum=;
     /*
     for(int i=0;i<10;i++)
         for(int j=0;j<10;j++)
             printf(j==9?"%d\n":"%d ",ans.g[i][j]);
     */
     for(int i=;i<;i++){
         sum+=ans.g[][i]*i;
         sum%=m;
     }
     printf("%d\n",sum);
 }
 int main(void)
 {
     int k;
     while(scanf("%d%d",&k,&m)!=EOF)
     {
         for(int i=;i<;i++)
             for(int j=;j<;j++)
                 temp.g[i][j]=ans.g[i][j]=;
         for(int i=;i<;i++)
             scanf("%d",&g[i]);
         for(int i=;i<;i++){
             if(i<) temp.g[i][i+]=;
             ans.g[i][i]=;
             temp.g[][i]=g[-i];
         }
         solve(k-);
     }
     return ;
 }