[title link]
[main ideas]
- The main chain of the caterpillar is defined as the part of the node with the degree of removal of 111.
- If the length of the main chain is less than 222, it is a chrysanthemum diagram, obviously there are N − [N=2]N-[N=2]N − [N=2] kinds.
- Consider enumerating the length of the main chain i (i ≥ 2)i\ (i\geq 2)i (i ≥ 2), arranging the points on the main chain arbitrarily, there are Ni2 \ frac {n ^ {i} {2} 2Ni.
- For the remaining nodes, we require that the head and tail of the main chain must have nodes, and whether the remaining nodes have any children.
- According to the inclusion exclusion principle, the number of hung sub nodes is iN − I − 2(i − 1)N − i+(i − 2)N − ii^{N-i}-2(i-1)^{N-i}+(i-2)^{N-i}iN − I − 2(i − 1)N − i+(i − 2)N − I.
- Time complexity O(NLogN)O(NLogN)O(NLogN).
- Using the method of generating function, all the answers of 1 ∼ N1\sim N1 ∼ n can be obtained.
[Code]
#include<bits/stdc++.h> using namespace std; const int MAXN = 2e5 + 5; const int P = 998244353; typedef long long ll; typedef long double ld; typedef unsigned long long ull; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } int fac[MAXN], inv[MAXN]; int power(int x, int y) { if (y == 0) return 1; int tmp = power(x, y / 2); if (y % 2 == 0) return 1ll * tmp * tmp % P; else return 1ll * tmp * tmp % P * x % P; } int getc(int x, int y) { if (y > x) return 0; else return 1ll * fac[x] * inv[y] % P * inv[x - y] % P; } void init(int n) { fac[0] = 1; for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % P; inv[n] = power(fac[n], P - 2); for (int i = n - 1; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1ll) % P; } void update(int &x, int y) { x += y; if (x >= P) x -= P; } int main() { int n; read(n); init(n); int ans = n - (n == 2); for (int i = 2; i <= n - 2; i++) { int coef = 1ll * fac[n] * inv[n - i] % P * inv[2] % P; int now = power(i, n - i); update(now, P - 2ll * power(i - 1, n - i) % P); update(now, power(i - 2, n - i)); update(ans, 1ll * now * coef % P); } writeln(ans); return 0; }