2019 ICPC南昌 J.Summon

发表于 | 阅读数

在环上涂颜色，不能出现给定颜色的顺时针顺序。（长度为=4)。

变换一下就是

然后$f(d)$就表示，长度$d$的环符合 要求有几个（不用考虑选择）
刚开始觉得这题，用快速幂挺炫酷的，发现就是套路。转化成从某个状态走$d$步到某个状态的步数种类。

跟$POJ2888$一摸一样。

变成四进制。然后矩阵快速幂没了。

  • 注意预处理可以转移的进制。
  • 常数优化。
代码 
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#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 1e3 + 10;
const int mod = 998244353;

const int Z = 64;
struct Matrix
{

    int row, col;
    int s[Z][Z];
    Matrix()
    { //无参数构造函数
        row = col = 0;
        memset(s, 0, sizeof(s));
    }
    Matrix(int n, int m, int x = 0) : row(n), col(m)
    {
        memset(s, 0, sizeof(s));
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                s[i][j] = x;
    } //重载构造函数

    void printp()
    { //输出一个矩阵
        for (int i = 0; i < row; i++)
        {
            for (int j = 0; j < col; j++)
                cout << s[i][j] << " ";
            cout << endl;
        }
    }
    // void transpose()
    // { //矩阵的转置
    //     Matrix y(col, row);
    //     for (int i = 1; i <= row; i++)
    //     {
    //         for (int j = 1; j <= col; j++)
    //             y.s[j][i] = s[i][j];
    //     }
    //     *this = y;
    // }
    void identity()
    { //将该矩阵改成单位矩阵
        for (int i = 0; i < row; i++)
            s[i][i] = 1;
    }
    // Matrix operator+(Matrix x)
    // { //重载+
    //     Matrix y(row, col);
    //     for (int i = 0; i <row; i++)
    //     {
    //         for (int j = 0; j < col; j++)
    //             y.s[i][j] += s[i][j] + x.s[i][j], y.s[i][j] %= mod;
    //     }
    //     return y;
    // }
    Matrix operator*(Matrix x)
    { //重载*
        Matrix y(row, x.col);
        for (int i = 0; i < row; i++)
        {
            for (int j = 0; j < x.col; j++)
            {
                for (int k = 0; k < col; k++)
                {
                    if (!s[i][k] || !x.s[k][j])
                        continue;
                    y.s[i][j] = (y.s[i][j] + 1ll * s[i][k] * x.s[k][j] % mod) % mod;
                }
            }
        }
        return y;
    }
    Matrix operator*=(Matrix x)
    { //重载*=
        return *this = *this * x;
    }
} z[40];
int qpow(int a, int x, int mo)
{
    int res = 1;
    while (x)
    {
        if (x & 1)
            res = 1ll * res * a % mod;
        a = 1ll * a * a % mod;
        x >>= 1;
    }
    return res;
}

Matrix qpow(int x)
{
    Matrix res(z[0].row, z[0].col);
    res.identity();

    for (int i = 0; (1 << i) <= x; i++)
    {
        if ((1 << i) & x)
        {
            res = res * z[i];
        }
    }

    return res;
}

namespace Polya
{

    int mult(int x, int y)
    {
        return 1ll * x * y % mod;
    }
    int calc(int x)
    {
        //  dt.printp();
        Matrix res = qpow(x);

        int ans = 0;
        for (int i = 0; i < 64; i++)
            ans = (ans + res.s[i][i]) % mod;

        return ans;
    }
    int p[N], e[N];
    int cnt = 0;
    int ans = 0;
    void dfs(int x, int now, int phi, int n)
    {
        if (x > cnt)
        {
            ans = (ans + mult(phi % mod, calc(n / now))) % mod;

            return;
        }
        dfs(x + 1, now, phi, n);
        for (int i = 1; i <= e[x]; i++)
        {
            phi *= (p[x] - (i == 1));
            now *= p[x];
            dfs(x + 1, now, phi, n);
        }
    }
    int solve(int n)
    {
        cnt = 0;
        ans = 0;
        int tmp = n;
        for (int i = 2; i * i <= n; i++)
            if (n % i == 0)
            {
                p[++cnt] = i;
                e[cnt] = 0;
                while (n % i == 0)
                {
                    n /= i;
                    e[cnt]++;
                }
            }
        if (n > 1)
        {
            p[++cnt] = n;
            e[cnt] = 1;
        }
        dfs(1, 1, 1, tmp);
        //cout << ans << endl;
        return 1ll * ans * qpow(tmp % mod, mod - 2, mod) % mod;
    }
} // namespace Polya

int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}

int main()
{
    int n = read(), m = read();
    Matrix tmp(64, 64, 0);
    for (int i = 0; i < 64; i++)
        for (int j = 0; j < 64; j++)
        {
            if (((j << 2) & 63) == ((i >> 2) << 2))
                tmp.s[i][j] = 1;
        }
    for (int i = 1; i <= m; i++)
    {
        vector<int> a(4);
        for (int j = 0; j < 4; j++)
        {
            a[j] = read();
        }
        tmp.s[a[0] + (a[1] << 2) + (a[2] << 4)][a[1] + (a[2] << 2) + (a[3] << 4)] = 0;
    }
    z[0] = tmp;
    for (int i = 1; (1 << i) <= n; i++)
        z[i] = z[i - 1] * z[i - 1];

    cout << Polya::solve(n) << endl;
}