HDU 5552 CDQ分治+NTT

题意：

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5552
统计n个点可以组成多少种不同的有环连通图，其中每条边可以染成m种颜色。

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思路：

分别统计f(n)为n个点的连通图数目，g(n)为n个点能构成的图数目，h(n)为n个点能构成的树的数目，然后递推出公式，发现是个卷积形式，可以用FFT加速，这里会爆精度，改用NTT。

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代码：

#include <bits/stdc++.h>using namespace std;typedef long long LL;const LL MOD = 152076289;const int MAXN = 4e4 + 10;const int G = 106;LL x1[MAXN], x2[MAXN];LL pow_mod(LL a, LL n) {    LL res = 1;    while (n) {        if (n & 1) res = res * a % MOD;        a = a * a % MOD;        n >>= 1;    }    return res;}void change (LL *y, int len) {    int i, j, k;    for(i = 1, j = len / 2; i < len - 1; i++) {        if (i < j) swap(y[i], y[j]);        k = len / 2;        while (j >= k) {            j -= k;            k /= 2;        }        if (j < k) j += k;    }}void ntt (LL *y, int len, int on) {    change (y, len);    int id = 0;    for(int h = 2; h <= len; h <<= 1) {        id++;        LL wn = pow_mod (G, (MOD - 1) / (1<<id));        for(int j = 0; j < len; j += h) {            LL w = 1;            for(int k = j; k < j + h / 2; k++) {                LL u = y[k] % MOD;                LL t = w * (y[k + h / 2] % MOD) % MOD;                y[k] = (u + t) % MOD;                y[k + h / 2] = ((u - t) % MOD + MOD) % MOD;                w = w * wn % MOD;            }        }    }    if (on == -1) {        for (int i = 1; i < len / 2; i++)            swap (y[i], y[len - i]);        LL inv = pow_mod(len, MOD - 2);        for(int i = 0; i < len; i++)            y[i] = y[i] % MOD * inv % MOD;    }}LL f[MAXN], g[MAXN], h[MAXN];LL inv[MAXN], fac[MAXN];void init() {    fac[0] = 1; inv[0] = 1;    for (int i = 1; i <= 40000; i++) {        fac[i] = fac[i - 1] * i % MOD;        inv[i] = pow_mod(fac[i], MOD - 2);    }}void solve(int l, int r) {    if (l == r) {        f[l] = (f[l] + g[l]) % MOD;        return;    }    int m = (l + r) >> 1;    solve(l, m);    int len = 1;    while (len <= r - l + 1) len <<= 1;    for (int i = 0; i < len; i++) x1[i] = x2[i] = 0;    for (int i = l; i <= m; i++) x1[i - l] = f[i] * inv[i - 1] % MOD;    for (int i = 1; i <= r - l; i++) x2[i - 1] = g[i] * inv[i] % MOD;    ntt(x1, len, 1); ntt(x2, len, 1);    for (int i = 0; i < len; i++) x1[i] = x1[i] * x2[i] % MOD;    ntt(x1, len, -1);    for (int i = m + 1; i <= r; i++) {        f[i] -= x1[i - l - 1] % MOD * fac[i - 1] % MOD;        f[i] = (f[i] + MOD) % MOD;    }    solve(m + 1, r);}int main() {    //freopen("in.txt", "r", stdin);    init();    int T, cs = 0;    scanf("%d", &T);    while (T--) {        int n, m;        scanf("%d%d", &n, &m);        for (int i = 1; i <= n; i++) g[i] = pow_mod(m + 1, i * (i - 1) / 2);        memset(f, 0, sizeof(f));        solve(1, n);        LL ans = f[n];        ans -= pow_mod(n, n - 2) * pow_mod(m, n - 1) % MOD;        ans = (ans + MOD) % MOD;        printf("Case #%d: %I64d\n", ++cs, ans);    }    return 0;}