HDU 5896 CDQ分治+NTT

题意：

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5896
求n个点能构成的有环图的数目。

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思路：

母题是HDU 5552：http://blog.csdn.net/Bahuia/article/details/78109301
两题思路一致，只是改变了所求结果，推导发生变化。

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代码：

#include <bits/stdc++.h>using namespace std;typedef long long LL;const LL MOD = 998244353;const int MAXN = 5e5 + 10;const int G = 3;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 g[MAXN], h[MAXN];LL inv[MAXN], fac[MAXN], two[MAXN], itwo[MAXN];void init() {    fac[0] = 1; inv[0] = 1, two[0] = 1, itwo[0] = 1;    for (int i = 1; i <= 100000; i++) {        fac[i] = fac[i - 1] * i % MOD;        two[i] = two[i - 1] * 2 % MOD;    }    inv[100000] = pow_mod(fac[100000], MOD - 2);    itwo[100000] = pow_mod(two[100000], MOD - 2);    for (int i = 100000 - 1; i >= 1; i--) {        inv[i] = inv[i + 1] * (i + 1) % MOD;        itwo[i] = itwo[i + 1] * 2 % MOD;    }}int n;LL ans[MAXN], a[MAXN];void solve() {    for (int i = 1; i <= n; i++) {        g[i] = fac[i - 1] * a[i] % MOD;        h[i] = two[i] * inv[n - i] % MOD;    }    int len = 1;    while (len <= 2 * n) len <<= 1;    x1[0] = x2[0] = 0;    for (int i = 1; i <= n; i++) x1[i] = g[i], x2[i] = h[i];    for (int i = n + 1; i <= len; i++) x1[i] = x2[i] = 0;    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 = 1; i <= n; i++)        ans[i] = x1[i + n] * itwo[i] % MOD * inv[i - 1] % MOD;    for (int i = 1; i <= n; i++) {        ans[i] += ans[i - 1];        ans[i] %= MOD;    }    for (int i = 1; i <= n; i++)        printf("%I64d ", ans[i]);    puts("");}bool cmp(const LL &x, const LL &y) {    return x > y;}int main() {    //freopen("in.txt", "r", stdin);    init();    int T;    scanf("%d", &T);    while (T--) {        scanf("%d", &n);        for (int i = 1; i <= n; i++)            scanf("%I64d", &a[i]);        sort(a + 1, a + 1 + n, cmp);        solve();    }    return 0;}