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=0;i<10;i++)
        for(int j=0;j<10;j++){
            c.g[i][j]=0;
            for(int k=0;k<10;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&1) ans=mul(temp,ans);
        temp=mul(temp,temp);
        k/=2;
    }
    int sum=0;
    /*
    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=0;i<10;i++){
        sum+=ans.g[9][i]*i;
        sum%=m;
    }
    printf("%d\n",sum);
}
int main(void)
{
    int k;
    while(scanf("%d%d",&k,&m)!=EOF)
    {
        for(int i=0;i<10;i++)
            for(int j=0;j<10;j++)  
                temp.g[i][j]=ans.g[i][j]=0;
        for(int i=0;i<10;i++)
            scanf("%d",&g[i]);
        for(int i=0;i<10;i++){
            if(i<9) temp.g[i][i+1]=1;
            ans.g[i][i]=1;
            temp.g[9][i]=g[9-i];
        }
        solve(k-9);
    }
    return 0;
}