宝宝眼屎多,陈丽华前夫离婚原因,上海婚假
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一开始想的是直接设\(f_i\)表示\(i\)个点的无向联通图个数，枚举最后一个联通块转移，发现有一种情况转移不到。。。
正解是先设\(g(n)\)表示\(n\)个点的无向图个数，这个方案是\(2^{\frac{i(i-1)}{2}}\)(也就是考虑每条边选不选)
考虑如何得到\(g\)
\[g(n) = \sum_{i=0}^n c_{n-1}^{i-1}f(i) g(n-i)\]
直接将\(2^{\frac{n(n-1)}{2}}\)带入然后化简一下可以得到这个式子
\[\frac{2^{c_n^2}}{(n-1)!} = \sum_{i=1}^n \frac{f(i)}{(i-1)!} \frac{2^{c_{n-i}^2}}{(n-i)!}\]
然后就可以多项式求逆啦。
我要评论#include<bits/stdc++.h>
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
#define ll long long
#define ull unsigned long long
#define fin(x) {freopen(#x".in","r",stdin);}
#define fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int maxn = 2e6 + 10, inf = 1e9 + 1;
const double eps = 1e-9, pi = acos(-1);
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int n, m, a[maxn], b[maxn], c[maxn], d[maxn], fac[maxn], ifac[maxn];
namespace poly {
int rev[maxn], gpow[maxn], a[maxn], b[maxn], c[maxn], lim, inv2;
const int g = 3, mod = 1004535809, mod2 = 1004535808;
template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;}
template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;}
int fp(int a, int p, int p = mod) {
int base = 1;
for(; p > 0; p >>= 1, a = 1ll * a * a % p) if(p & 1) base = 1ll * base * a % p;
return base;
}
int inv(int x) {
return fp(x, mod - 2);
}
int getlen(int x) {
int lim = 1;
while(lim <= x) lim <<= 1;
return lim;
}
int getorigin(int x) {//¼æëãô¸ù
static int q[maxn]; int tot = 0, tp = x - 1;
for(int i = 2; i * i <= tp; i++) if(!(tp % i)) {q[++tot] = i;while(!(tp % i)) tp /= i;}
if(tp > 1) q[++tot] = tp;
for(int i = 2, j; i <= x - 1; i++) {
for(j = 1; j <= tot; j++) if(fp(i, (x - 1) / q[j], x) == 1) break;
if(j == tot + 1) return i;
}
return -1;
}
void init(/*int p,*/ int lim) {
inv2 = fp(2, mod - 2);
for(int i = 1; i <= lim; i++) gpow[i] = fp(g, (mod - 1) / i);
}
void ntt(int *a, int lim, int opt) {
int len = 0; for(int n = 1; n < lim; n <<= 1) ++len;
for(int i = 1; i <= lim; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (len - 1));
for(int i = 0; i <= lim; i++) if(i < rev[i]) swap(a[i], a[rev[i]]);
for(int mid = 1; mid < lim; mid <<= 1) {
int wn = gpow[mid << 1];
for(int i = 0; i < lim; i += (mid << 1)) {
for(int j = 0, w = 1; j < mid; j++, w = mul(w, wn)) {
int x = a[i + j], y = mul(w, a[i + j + mid]);
a[i + j] = add(x, y), a[i + j + mid] = add(x, -y);
}
}
}
if(opt == -1) {
reverse(a + 1, a + lim);
int inv = fp(lim, mod - 2);
for(int i = 0; i <= lim; i++) mul2(a[i], inv);
}
}
void mul(int *a, int *b, int n, int m) {
memset(a, 0, sizeof(a)); memset(b, 0, sizeof(b));
int lim = 1, len = 0;
while(lim <= n + m) len++, lim <<= 1;
for(int i = 0; i <= n; i++) a[i] = a[i];
for(int i = 0; i <= m; i++) b[i] = b[i];
ntt(a, lim, 1); ntt(b, lim, 1);
for(int i = 0; i <= lim; i++) b[i] = mul(b[i], a[i]);
ntt(b, lim, -1);
for(int i = 0; i <= n + m; i++) b[i] = b[i];
memset(a, 0, sizeof(a)); memset(b, 0, sizeof(b));
}
void inv(int *a, int *b, int len) {//b1 = 2b - a1 * b^2
if(len == 1) {b[0] = fp(a[0], mod - 2); return ;}
inv(a, b, len >> 1);
for(int i = 0; i < len; i++) a[i] = a[i], b[i] = b[i];
ntt(a, len << 1, 1); ntt(b, len << 1, 1);
for(int i = 0; i < (len << 1); i++) mul2(a[i], mul(b[i], b[i]));
ntt(a, len << 1, -1);
for(int i = 0; i < len; i++) add2(b[i], add(b[i], -a[i]));
for(int i = 0; i < (len << 1); i++) a[i] = b[i] = 0;
}
void dao(int *a, int *b, int len) {
for(int i = 1; i < len; i++) b[i - 1] = mul(i, a[i]); b[len - 1] = 0;
}
void ji(int *a, int *b, int len) {
for(int i = 1; i < len; i++) b[i] = mul(a[i - 1], fp(i, mod - 2)); b[0] = 0;
}
void ln(int *a, int *b, int len) {//g(a) = \frac{a}{a'} qiudao zhihou jifen
static int a[maxn], b[maxn];
dao(a, a, len);
inv(a, b, len);
ntt(a, len << 1, 1); ntt(b, len << 1, 1);
for(int i = 0; i < (len << 1); i++) b[i] = mul(a[i], b[i]);
ntt(b, len << 1, -1);
ji(b, b, len << 1);
memset(a, 0, sizeof(a)); memset(b, 0, sizeof(b));
}
void exp(int *a, int *b, int len) {//f(x) = f_0 (1 - lnf_0 + a) but code ..why....
if(len == 1) return (void) (b[0] = 1);
exp(a, b, len >> 1); ln(b, c, len);
c[0] = add(a[0] + 1, -c[0]);
for(int i = 1; i < len; i++) c[i] = add(a[i], -c[i]);
ntt(c, len << 1, 1); ntt(b, len << 1, 1);
for(int i = 0; i < (len << 1); i++) mul2(b[i], c[i]);
ntt(b, len << 1, -1);
for(int i = len; i < (len << 1); i++) c[i] = b[i] = 0;
}
void sqrt(int *a, int *b, int len) {
static int b[maxn];
ln(a, b, len);
for(int i = 0; i < len; i++) b[i] = mul(b[i], inv2);
exp(b, b, len);
}
};
using namespace poly;
signed main() {
n = read(); int lim = getlen(n); init(4 * lim);
fac[0] = 1;
for(int i = 1; i <= n; i++) fac[i] = mul(i, fac[i - 1]);
ifac[n] = fp(fac[n], mod - 2);
for(int i = n; i >= 1; i--) ifac[i - 1] = mul(i, ifac[i]);
for(int i = n; i >= 1; i--) {
int tmp;
if(i & 1) tmp = fp(2, 1ll * ((i - 1) / 2) * i % mod2);
else tmp = fp(2, 1ll * (i / 2) * (i - 1) % mod2);
a[i] = mul(tmp, ifac[i - 1]),
b[i] = mul(tmp, ifac[i]);
}
b[0] = 1;
inv(b, c, lim);
mul(a, c, lim, lim);
cout << mul(c[n], fac[n - 1]);
return 0;
}
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