HDU 6030 Happy Necklace

矩阵快速幂。

因为任意素数长度都要满足，所以$3$必须满足，$3$一旦满足，其余的肯定满足，也就是说只要考虑字符串末尾两位即可，$dp$一下就可以算方案数了。$n$较大，可以矩阵加速。

#include <bits/stdc++.h>
using namespace std;
const long long mod=1e9+;
const long long inf=1e18;

int T;
long long n;

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

Matrix X, Y, Z;

Matrix Matrix::operator*(Matrix b)
{
    Matrix c;
    memset(c.A, , sizeof(c.A));
    int i, j, k;
    for (i = ; i <= R; i++)
        for (j = ; j <= C; j++)
            for (k = ; 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()
{
    n = n - ;
    memset(X.A,,sizeof X.A);
    memset(Y.A,,sizeof Y.A);
    memset(Z.A,,sizeof Z.A);

    Z.R = ; Z.C = ;
    Z.A[][]=; Z.A[][]=; Z.A[][]=;

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

    X.A[][]=; X.A[][]=; X.A[][]=;
    X.A[][]=; X.A[][]=; X.A[][]=;
    X.A[][]=; X.A[][]=; X.A[][]=;

    X.R = ; X.C = ;
}

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

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

int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld",&n);
        init();
        work();
    }
    return ;
}