hdu 1757 A Simple Math Problem

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4038    Accepted Submission(s): 2437

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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典型的矩阵快速幂问题，第一次写，参考了别人的代码，感觉收益良多，最重要的地方在于构建一个矩阵。

 

借用一下别人代码的推论：

下面我们研究一下这道题如何运用矩阵。

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|

*/

我们看到规律了，每次要到下次个A*B，以此类推则由A*A*A.......A*B；

题意：按照题目给的公式，求f(k) % m的值。

附上代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
struct mat
{
    int m[10][10];
};
int mod;

mat mul(mat a,mat b)
{
    mat c;
    int i,j,k;
    memset(c.m,0,sizeof(c.m));
    for(i=0; i<10; i++)
        for(j=0; j<10; j++)
        {
            for(k=0; k<10; k++)
            {
                c.m[i][j]+=(a.m[i][k]*b.m[k][j])%mod;
            }
            c.m[i][j]%=mod;
        }
    return c;
}

mat product(mat a,int k)
{
    if(k==1) return a;
    else if(k&1)
        return mul(product(a,k-1),a);
    else
        return product(mul(a,a),k/2);
}

int main()
{
    int i,j,k;
    mat a;
    while(~scanf("%d%d",&k,&mod))
    {
        for(i=9; i>=0; i--)
            scanf("%d",&a.m[9][i]);
        if(k<10)
        {
            printf("%d
",k%mod);
            continue;
        }
        for(i=0; i<9; i++)
            for(j=0; j<10; j++)
                if(j==i+1) a.m[i][j]=1;
                else a.m[i][j]=0;
        mat b=product(a,k-9);
        int ans=0;
        for(i=0; i<10; i++)
            ans+=(b.m[9][i]*i)%mod;
        printf("%d
",ans%mod);
    }
    return 0;
}