点击此处即可传送到hdu 1757

                        **A Simple Math Problem**
Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3497    Accepted Submission(s): 2112



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

题目大意：给你两个数,k,mod,
然后再有是个数a[i],就是求 f(k) % mod
解题思路：很简单，就是用一下那个矩阵乘法和快速幂就行了，就是套模板，具体详见我代码：

上代码：

/*
2015 - 8 - 15 下午
Author: ITAK

今天感觉还可以，买了一个机械键盘，又要少吃几顿饭了。。。

今日的我要超越昨日的我，明日的我要胜过今日的我，
以创作出更好的代码为目标，不断地超越自己。
*/
#include <iostream>
#include <cstdio>
using namespace std;
const int maxn = 10;
int mod;
typedef long long LL;
typedef  struct
{
    int  m[maxn][maxn];
}  Matrix;
//定义一个10*10的矩阵
Matrix P= {0,0,0,0,0,0,0,0,0,0,
           1,0,0,0,0,0,0,0,0,0,
           0,1,0,0,0,0,0,0,0,0,
           0,0,1,0,0,0,0,0,0,0,
           0,0,0,1,0,0,0,0,0,0,
           0,0,0,0,1,0,0,0,0,0,
           0,0,0,0,0,1,0,0,0,0,
           0,0,0,0,0,0,1,0,0,0,
           0,0,0,0,0,0,0,1,0,0,
           0,0,0,0,0,0,0,0,1,0
          };
Matrix I= {1,0,0,0,0,0,0,0,0,0,
           0,1,0,0,0,0,0,0,0,0,
           0,0,1,0,0,0,0,0,0,0,
           0,0,0,1,0,0,0,0,0,0,
           0,0,0,0,1,0,0,0,0,0,
           0,0,0,0,0,1,0,0,0,0,
           0,0,0,0,0,0,1,0,0,0,
           0,0,0,0,0,0,0,1,0,0,
           0,0,0,0,0,0,0,0,1,0,
           0,0,0,0,0,0,0,0,0,1
          };
//矩阵乘法
Matrix matrix_mul(Matrix a, Matrix b)
{
    int i,j,k;
    Matrix c;
    for(i=0; i<maxn; i++)
    {
        for(j=0; j<maxn; j++)
        {
            c.m[i][j] = 0;
            for(k=0; k<maxn; k++)
                c.m[i][j] += (a.m[i][k]*b.m[k][j])%mod;
            c.m[i][j] %= mod;//及时取余
        }
    }
    return c;
}
//矩阵的快速幂
Matrix quick_mod(LL m)
{
    Matrix ans = I, b = P;
    while(m)
    {
        if(m & 1)
            ans = matrix_mul(ans,b);
        m >>= 1;
        b = matrix_mul(b, b) ;
    }
    return ans;
}
int a[10];
int main()
{
    int k;
    while(~scanf("%d%d",&k,&mod))
    {
        for(int i=0; i<10; i++)
        {
            scanf("%d",&a[i]);
            P.m[0][i] = a[i];//给矩阵赋值
        }
        if(k < 10)
            printf("%d\n",k%mod);
        else
        {
            int sum = 0;
            Matrix tmp = quick_mod(k-9);
            for(int j=9; j>=0; j--)
                sum = (sum+tmp.m[0][9-j]*j)%mod;//不要最后加一边加，一边取余
            cout<<sum<<endl;
        }
    }
    return 0;
}