HDU 5015 233 Matrix

矩阵快速幂。

首先得到这个式子：

依据这个，就可以构造矩阵。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;

long long const MOD = 10000007;
int n, m;
long long a[20];

struct Matrix
{
    long long A[20][20];
    int R, C;
    Matrix operator*(Matrix b);
};

Matrix X, Y, Z;

Matrix Matrix::operator*(Matrix b)
{
    Matrix c;
    memset(c.A, 0, sizeof(c.A));
    int i, j, k;
    for (i = 1; i <= R; i++)
        for (j = 1; j <= b.C; j++)
            for (k = 1; k <= C; k++)
                c.A[i][j] = (c.A[i][j] + (A[i][k] * b.A[k][j]) % MOD) % MOD;
    c.R = R; c.C = b.C;
    return c;
}

void init()
{
    memset(X.A, 0, sizeof X.A);
    memset(Y.A, 0, sizeof Y.A);
    memset(Z.A, 0, sizeof Z.A);

    Z.R = n + 2; Z.C = 1;
    for (int i = 1; i <= n; i++) Z.A[i][1] = a[i]; Z.A[n + 1][1] = 23; Z.A[n + 2][1] = 3;

    X.R = n + 2; X.C = n + 2;
    for (int i = 1; i <= n; i++) for (int j = 1; j <= i; j++) X.A[i][j] = 1;
    for (int i = 1; i <= n + 1; i++) X.A[i][n + 1] = 10;
    for (int i = 1; i <= n + 2; i++) X.A[i][n + 2] = 1;

    Y.R = n + 2; Y.C = n + 2;
    for (int i = 1; i <= n + 2; i++) Y.A[i][i] = 1;
}

void read()
{
    for (int i = 1; i <= n; i++)
    {
        scanf("%lld", &a[i]);
        a[i] = a[i] % MOD;
    }
}

void work()
{
    while (m)
    {
        if (m % 2 == 1) Y = Y*X;
        m = m >> 1;
        X = X*X;
    }
    Z = Y*Z;

    printf("%lld\n", Z.A[n][1]);
}

int main()
{
    while (~scanf("%d%d", &n, &m))
    {
        read();
        init();
        work();
    }
    return 0;
}