UVa 10870 - Recurrences 矩阵快速幂

构造矩阵.

见《训练指南》p156...

#include <cstdio>
#include <cstring>
#include <cstdlib>

#define LL long long int

using namespace std;

const int MAXN = 20;

LL mod;

struct Matrix
{
    LL a[MAXN][MAXN];
    int r, c;
    friend Matrix operator*( Matrix &a, Matrix &b );
    friend Matrix operator^( Matrix a, LL k );
};

Matrix operator*( Matrix &a, Matrix &b )
{
    Matrix tmp;
    tmp.r = a.r;
    tmp.c = b.c;
    for ( int i = 1; i <= a.r; ++i )
        for ( int j = 1; j <= b.c; ++j )
        {
            tmp.a[i][j] = 0;
            for ( int k = 1; k <= a.c; ++k )
            {
                //printf( "%d %d %d\n", i, j, k );
                tmp.a[i][j] = ( tmp.a[i][j] + (a.a[i][k] * b.a[k][j]) % mod ) % mod;
            }
        }
    return tmp;
}

Matrix operator^( Matrix a, LL k )
{
    Matrix unit;
    unit.r = a.r;
    unit.c = a.c;
    memset( unit.a, 0, sizeof(unit.a) );
    for ( int i = 0; i < MAXN; ++i )
        unit.a[i][i] = 1;

    while ( k )
    {
        if ( k & 1 ) unit = unit * a;
        a = a * a;
        k = ( k >> 1 );
    }
    return unit;
}

/**************以上矩阵快速幂模板*****************/

int n, d;
Matrix A, F;

int main()
{
    //freopen( "in.txt", "r", stdin );
    while ( scanf( "%d%d%lld", &d, &n, &mod ), d || n || mod )
    {
        memset( A.a, 0, sizeof(A.a) );
        for ( int i = 1; i < d; ++i )
            A.a[i][i + 1] = 1;

        for ( int i = d; i > 0; --i )
        {
            scanf( "%lld", &A.a[d][i] );
            A.a[d][i] %= mod;
        }
        A.r = d, A.c = d;

        memset( F.a, 0, sizeof(F.a) );
        for ( int i = 1; i <= d; ++i )
        {
            scanf( "%lld", &F.a[i][1] );
            F.a[i][1] %= mod;
        }
        F.r = d, F.c = 1;

        A = A ^ (n - d);
        F = A * F;
        printf( "%lld\n", F.a[d][1] % mod );

    }
    return 0;
}