結果
問題 | No.1507 Road Blocked |
ユーザー |
|
提出日時 | 2021-12-19 14:38:25 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 101 ms / 2,000 ms |
コード長 | 2,451 bytes |
コンパイル時間 | 4,021 ms |
コンパイル使用メモリ | 240,364 KB |
実行使用メモリ | 15,232 KB |
最終ジャッジ日時 | 2024-09-15 14:33:19 |
合計ジャッジ時間 | 9,361 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 30 |
ソースコード
#include <bits/stdc++.h>#include <atcoder/all>typedef long long int ll;typedef long long int ull;#define MP make_pairusing namespace std;using namespace atcoder;typedef pair<ll, ll> P;const ll MOD = 998244353;// const ll MOD = 1000000007;using mint = modint998244353;// using mint = modint1000000007;const double pi = 3.1415926536;const int MAX = 2000003;long long fac[MAX], finv[MAX], inv[MAX];// テーブルを作る前処理void COMinit() {fac[0] = fac[1] = 1;finv[0] = finv[1] = 1;inv[1] = 1;for (int i = 2; i < MAX; i++){fac[i] = fac[i - 1] * i % MOD;inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;finv[i] = finv[i - 1] * inv[i] % MOD;}}// 二項係数計算long long COM(int n, int k){if (n < k) return 0;if (n < 0 || k < 0) return 0;return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;}void comp(vector<ll>&a){set<ll>s(a.begin(),a.end());map<ll,ll>d;ll cnt=0;for(auto x:s)d[x]=cnt++;for(auto&x:a)x=d[x];}ll gcd (ll x, ll y) {if (y == 0) return x;else if (y > x) {return gcd (y, x);}else return gcd(x % y, y);}ll lcm (ll x, ll y) {return x / gcd(x, y) * y;}ll my_sqrt(ll x) {ll m = 0;ll M = 1000000001;while (M - m > 1) {ll now = (M + m) / 2;if (now * now <= x) {m = now;}else {M = now;}}return m;}ll keta(ll n) {ll ret = 0;while (n) {n /= 10;ret++;}return ret;}ll ceil(ll n, ll m) {ll ret = n / m;if (n % m) ret++;return ret;}ll pow_ll(ll x, ll n) {if (n == 0) return 1;if (n % 2) {return pow_ll(x, n - 1) * x;}else {ll tmp = pow_ll(x, n / 2);return tmp * tmp;}}vector<ll> e[100001];ll n;mint p = 0;ll dfs(int cur, int par) {ll ret = 1;for (auto nex : e[cur]) {if (nex == par) continue;ll tmp = dfs(nex, cur);p += tmp * (n - tmp);ret += tmp;}return ret;}int main() {cin >> n;for (int i = 1; i <= n - 1; i++) {ll u, v;cin >> u >> v;e[u].push_back(v);e[v].push_back(u);}dfs(1, 0);mint q = (n - 1) * (n - 1) * n / 2;p = q - p;// cout << p.val() << ' ' << q.val() << endl;mint ans = p * pow_mod(q.val(), MOD - 2, MOD);cout << ans.val() << endl;return 0;}