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+7;
const long long inf=1e18;

int T;
long long n;

struct Matrix
{
    long long A[5][5];
    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 <= 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()
{
    n = n - 2;
    memset(X.A,0,sizeof X.A);
    memset(Y.A,0,sizeof Y.A);
    memset(Z.A,0,sizeof Z.A);

    Z.R = 1; Z.C = 3;
    Z.A[1][1]=1; Z.A[1][2]=1; Z.A[1][3]=1;

    for(int i=1;i<=3;i++) Y.A[i][i]=1; Y.R = 3; Y.C = 3;

    X.A[1][1]=1; X.A[1][2]=0; X.A[1][3]=1;
    X.A[2][1]=1; X.A[2][2]=0; X.A[2][3]=0;
    X.A[3][1]=0; X.A[3][2]=1; X.A[3][3]=0;

    X.R = 3; X.C = 3;
}

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

    printf("%lld\n", (Z.A[1][1]+Z.A[1][2]+Z.A[1][3])%mod);
}


int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld",&n);
        init();
        work();
    }
    return 0;
}