hdu1757---A Simple Math Problem(矩阵)

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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/*************************************************************************    > File Name: hdu1757.cpp    > Author: ALex    > Mail: [e-mail]     > Created Time: 2015年03月11日 星期三 19时57分20秒 ************************************************************************/#include <map>#include <set>#include <queue>#include <stack>#include <vector>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const double pi = acos(-1);const int inf = 0x3f3f3f3f;const double eps = 1e-15;typedef long long LL;typedef pair <int, int> PLL;int mod;class MARTIX{    public:        int mat[105][105];        MARTIX();        MARTIX operator + (const MARTIX &b)const;        MARTIX operator * (const MARTIX &b)const;        MARTIX & operator = (const MARTIX &b);};MARTIX :: MARTIX(){    memset (mat, 0, sizeof(mat));}MARTIX MARTIX :: operator + (const MARTIX &b)const{    MARTIX ret;    for (int i = 0; i < 10; ++i)    {        for (int j = 0; j < 10; ++j)        {            ret.mat[i][j] = mat[i][j] + b.mat[i][j];            ret.mat[i][j] %= mod;        }    }    return ret;}MARTIX MARTIX :: operator * (const MARTIX &b)const{    MARTIX ret;    for (int i = 0; i < 10; ++i)    {        for (int j = 0; j < 10; ++j)        {            ret.mat[i][j] = 0;            for (int k = 0; k < 10; ++k)            {                ret.mat[i][j] += mat[i][k] * b.mat[k][j];                ret.mat[i][j] %= mod;            }        }    }    return ret;}MARTIX & MARTIX :: operator = (const MARTIX  &b){    for (int i = 0; i < 10; ++i)    {        for (int j = 0; j < 10; ++j)        {            this -> mat[i][j] = b.mat[i][j];        }    }    return *this;}MARTIX fastpow(MARTIX A, LL n){    MARTIX ans;    for (int i = 0; i < 10; ++i)    {        for (int j = 0; j < 10; ++j)        {            ans.mat[i][j] = (i == j);        }    }    while (n)    {        if (n & 1)        {            ans = ans * A;        }        n >>= 1;        A = A * A;    }    return ans;}void Debug(MARTIX A){    for (int i = 0; i < 10; ++i)    {        for (int j = 0; j < 10; ++j)        {            printf("%d ", A.mat[i][j]);        }        printf("\n");    }}int main (){    LL k;    while (~scanf("%lld%d", &k, &mod))    {        if (k < 10)        {            printf("%lld\n", k);            continue;        }        MARTIX A;        for (int i = 0; i < 10; ++i)        {            scanf("%d", &A.mat[i][0]);        }        for (int i = 0; i < 9; ++i)        {            A.mat[i][i + 1] = 1;        }        MARTIX ans = fastpow(A, k - 9);        MARTIX F;        for (int i = 0; i < 10; ++i)        {            F.mat[0][i] = 10 - i - 1;        }//      Debug(F);        ans = F * ans;        printf("%d\n", ans.mat[0][0]);    }    return 0;}