結果
問題 | No.2360 Path to Integer |
ユーザー |
|
提出日時 | 2023-07-06 00:38:16 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 160 ms / 2,500 ms |
コード長 | 5,263 bytes |
コンパイル時間 | 3,763 ms |
コンパイル使用メモリ | 247,132 KB |
実行使用メモリ | 32,436 KB |
最終ジャッジ日時 | 2024-07-19 18:53:41 |
合計ジャッジ時間 | 5,771 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 15 |
コンパイルメッセージ
main.cpp: In function 'void dfs2(int, int, std::vector<std::__cxx11::basic_string<char> >&, std::vector<std::vector<int> >&, std::vector<std::vector<atcoder::static_modint<998244353> > >&, mint)': main.cpp:121:23: warning: 'parIdx' may be used uninitialized [-Wmaybe-uninitialized] 121 | dp[cur][parIdx] = 1 + psum; | ^ main.cpp:113:13: note: 'parIdx' was declared here 113 | int parIdx; | ^~~~~~ main.cpp: In function 'int dfs3(int, int, std::vector<std::vector<int> >&, std::vector<std::vector<int> >&)': main.cpp:153:24: warning: 'parIdx' may be used uninitialized [-Wmaybe-uninitialized] 153 | num[cur][parIdx] = edge.size() - ret; | ^ main.cpp:141:9: note: 'parIdx' was declared here 141 | int parIdx; | ^~~~~~
ソースコード
#include <bits/stdc++.h>#include <atcoder/all>typedef long long int ll;using namespace std;typedef pair<ll, ll> P;using namespace atcoder;template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;#define USE998244353#ifdef USE998244353const ll MOD = 998244353;// const double PI = 3.141592653589;using mint = modint998244353;#elseconst ll MOD = 1000000007;using mint = modint1000000007;#endifconst int MAX = 20000001;long long fac[MAX], finv[MAX], inv[MAX];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;}}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;}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 = 3000000001;while (M - m > 1) {ll now = (M + m) / 2;if (now * now <= x) {m = now;}else {M = now;}}return m;}ll keta(ll num, ll arity) {ll ret = 0;while (num) {num /= arity;ret++;}return ret;}ll ceil(ll n, ll m) {// n > 0, m > 0ll 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> compress(vector<ll>& v) {// [3 5 5 6 1 1 10 1] -> [1 2 2 3 0 0 4 0]vector<ll> u = v;sort(u.begin(), u.end());u.erase(unique(u.begin(),u.end()),u.end());map<ll, ll> mp;for (int i = 0; i < u.size(); i++) {mp[u[i]] = i;}for (int i = 0; i < v.size(); i++) {v[i] = mp[v[i]];}return v;}mint dfs1(int cur, int par, vector<string>& a, vector<vector<int>>& edge, vector<vector<mint>>& dp) {mint ret = 1;for (int i = 0; i < edge[cur].size(); i++) {int nex = edge[cur][i];if (nex == par) continue;mint tmp = dfs1(nex, cur, a, edge, dp);dp[cur][i] = tmp;ret += tmp;}ret *= pow_mod(10, a[cur].size(), MOD);return ret;}void dfs2(int cur, int par, vector<string>& a, vector<vector<int>>& edge, vector<vector<mint>>& dp, mint psum) {if (par != -1) {int parIdx;for (int i = 0; i < edge[cur].size(); i++) {int nex = edge[cur][i];if (nex == par) {parIdx = i;break;}}dp[cur][parIdx] = 1 + psum;dp[cur][parIdx] *= pow_mod(10, a[par].size(), MOD);}mint sm = 0;for (int i = 0; i < edge[cur].size(); i++) {sm += dp[cur][i];}for (int i = 0; i < edge[cur].size(); i++) {int nex = edge[cur][i];if (nex == par) {continue;}dfs2(nex, cur, a, edge, dp, sm - dp[cur][i]);}}int dfs3(int cur, int par, vector<vector<int>>& edge, vector<vector<int>>& num) {int ret = 1;int parIdx;for (int i = 0; i < edge[cur].size(); i++) {int nex = edge[cur][i];if (nex == par) {parIdx = i;continue;}int tmp = dfs3(nex, cur, edge, num);num[cur][i] = tmp;ret += tmp;}if (par != -1) {num[cur][parIdx] = edge.size() - ret;}return ret;}int main() {int n;cin >> n;vector<string> a(n);for (int i = 0; i < n; i++) cin >> a[i];vector<vector<int>> edge(n);for (int i = 0; i < n - 1; i++) {int u, v;cin >> u >> v;u--;v--;edge[u].push_back(v);edge[v].push_back(u);}vector<vector<mint>> dp(n);for (int i = 0; i < n; i++) {dp[i].resize(edge[i].size());}dfs1(0, -1, a, edge, dp);dfs2(0, -1, a, edge, dp, 0);vector<vector<int>> num(n);for (int i = 0; i < n; i++) {num[i].resize(edge[i].size());}dfs3(0, -1, edge, num);// dp[cur][i] ... 頂点curからi番目の辺を使った時に、全部でどれくらい掛け算されるか// num[cur][i]... 頂点curからi番目の辺を使った時の部分木の頂点の数// for (int i = 0; i < n; i++) {// for (int j = 0; j < edge[i].size(); j++) {// cout << i + 1 << ' ' << edge[i][j] + 1 << ' ' << num[i][j] << '\n';// }// }// mintmint ans = 0;for (int i = 0; i < n; i++) {mint nu = 0;for (auto c : a[i]) {nu *= 10;nu += (c - '0');}for (int j = 0; j < edge[i].size(); j++) {// cerr << i + 1 << ' ' << edge[i][j] + 1 << ' ' << dp[i][j].val() << ' ' << num[i][j] << '\n';ans += (dp[i][j]) * nu * (n - num[i][j]);}ans += n * nu;}cout << ans.val() << '\n';return 0;}