結果
| 問題 | No.2515 Similar Triangles |
| ユーザー |
 元祖のヨッシー
|
| 提出日時 | 2023-10-27 21:23:32 |
| 言語 | C++23 (gcc 12.3.0 + boost 1.83.0) |
| 結果 |
AC
|
| 実行時間 | 2 ms / 2,000 ms |
| コード長 | 25,556 bytes |
| コンパイル時間 | 15,005 ms |
| コンパイル使用メモリ | 589,504 KB |
| 実行使用メモリ | 4,348 KB |
| 最終ジャッジ日時 | 2023-10-27 21:23:48 |
| 合計ジャッジ時間 | 9,249 ms |
|
ジャッジサーバーID （参考情報） |
judge11 / judge12 |
（要ログイン）
テストケース
テストケース表示| 入力 | 結果 | 実行時間 実行使用メモリ |
|---|---|---|
| testcase_00 | AC | 2 ms
4,348 KB |
| testcase_01 | AC | 2 ms
4,348 KB |
| testcase_02 | AC | 2 ms
4,348 KB |
| testcase_03 | AC | 2 ms
4,348 KB |
| testcase_04 | AC | 2 ms
4,348 KB |
| testcase_05 | AC | 2 ms
4,348 KB |
| testcase_06 | AC | 1 ms
4,348 KB |
| testcase_07 | AC | 2 ms
4,348 KB |
| testcase_08 | AC | 2 ms
4,348 KB |
| testcase_09 | AC | 2 ms
4,348 KB |
ソースコード
#include <atcoder/all>/*
__________
< ずんだもんなのだ >
----------
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https://lunicode.com/imagetotext `'!!!!!!!'-` `!!(""""+!!!-*/
#include <bits/stdc++.h>
#include <random>
#include <boost/multiprecision/cpp_int.hpp>
#include <boost/multiprecision/cpp_dec_float.hpp>
//#include <random>
//#include <chrono>
#ifdef _DEBUG
#define _GLIBCXX_DEBUG
#endif
using namespace std;
//using namespace std::chrono;
using namespace atcoder;
using namespace boost::multiprecision;
#define rep(i, n) for (ll i = 0; i < (ll)(n); ++i)
#define repi(i,a) for (auto i = a.begin(); i!=a.end(); ++i)
#define repa(i, n) for (ll i = 1; i < (ll)(n); ++i)
#define reps(i,a, n) for (ll i = (ll)(a); i < (ll)(n); ++i)
#define repm(i,a, n) for (ll i = (ll)(a); i > (ll)(n); --i)
#define repbit(i, n) for (ll i = 0; i < (1LL<<n); ++i)
using ul = unsigned long long; using ll = long long; using ld = long double;
#define abs llabs
// 深さ優先探索
const ll INF = 1LL<<60; const long double PI = 3.14159265358979323;
const int INFi = INT_MAX;
struct Edge {
int to;
ll w;
Edge(int to, ll w) :to(to), w(w) {}
};
using Graph = vector<vector<Edge>>;
template<class T>bool chmin(T& a, T b) {
if (a > b) {
a = b; return true;//更新が発生した=trueを返す
}
return false;
}
template<class T>bool chmax(T& a, T b) {
if (a < b) {
a = b; return true;//更新が発生した=trueを返す
}
return false;
}
//返り値(負の閉路を持つかどうかを示すフラグ、最短路帳が格納されたvector)
pair<bool, vector<long long>> BellmanFord(Graph& G, int s) {
int N = (int)G.size();
bool exist_negative_cycle = false;//負の閉路を持つかどうか返す結果を初期化
vector<long long> dist(N, INF);
dist[s] = 0;
for (int i = 0; i < N; ++i) {
bool update = false; // 今回の(すべての)辺の緩和で更新が発⽣したかどうか
for (int v = 0; v < N; ++v) {
if (dist[v] == INF) continue;
for (auto e : G[v]) {
if (chmin(dist[e.to], dist[v] + e.w)) {
update = true;
}
}
}
if (!update) break; // 更新が発⽣しないならばすでに真の最短路⻑が求まっている
if (i == N - 1 && update) exist_negative_cycle = true; // 負の閉路がある場合
}
return pair<bool, vector<long long>>(exist_negative_cycle, dist);
}
vector<ll> Dijkstra(Graph& G, int s) {
int N = (int)G.size();
vector<long long> dist(N, INF);
// 要素が pair<long long, int>(= (最短路⻑, 路の終点の頂点の番号)) で、要素を昇順に並べたヒープを⽤意する
priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>>que;
dist[s] = 0;
que.emplace(0LL, s);
while (!que.empty()) {
auto [d, v] = que.top();
que.pop();
if (dist[v] < d)continue;
for (auto e : G[v]) {
if (chmin(dist[e.to], dist[v] + e.w)) {
que.emplace(dist[e.to], e.to);
}
}
}
return dist;
}
vector<int> BFS(vector<vector<int>>& G, int s) {
int N = (int)G.size();
vector<int> dist(N, -1); // dist[v] = -1 に初期化する
queue<int> que;
que.push(s);
dist[s] = 0; // 始点 s の最短路⻑は 0 に初期化する
while (!que.empty()) {
int v = que.front();
que.pop();
for (int i = 0; i < G[v].size(); i++) {
int w = G[v][i];
if (dist[w] != -1) continue;
dist[w] = dist[v] + 1; // 頂点 w の最短路⻑は (頂点 v の最短路⻑ + 1) になる
que.push(w);
}
}
return dist;
}
template <class T>
istream& operator>>(istream& is, vector<T>& v) {
for (auto& x : v)is >> x;
return is;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
for (int i = 0; i < (int)v.size(); i++)
if (i != (int)v.size() - 1)os << v[i] << endl; else os << v[i];
return os;
}
template <class T>
istream& operator>>(istream& is, set<T>& s) {
T cinset;
is >> cinset;
s.emplace(cinset);
return is;
}
template <class T>
istream& operator>>(istream& is, multiset<T>& s) {
T cinset;
is >> cinset;
s.emplace(cinset);
return is;
}
void topsort(vector<vector<int>>& G, int v, vector<bool>& seen, vector<int>& answer) {
seen[v] = true;
for (int i = 0; i < (int)G[v].size(); i++) {
int w = G[v][i];
if (seen[w]) continue;
topsort(G, w, seen, answer);
}
answer.emplace_back(v);
}
void dfs(vector<vector<int>>& G, int v, vector<bool>& seen) {
seen[v] = true;
for (int i = 0; i < (int)G[v].size(); i++) {
int w = G[v][i];
if (seen[w])continue;
dfs(G, w, seen);
}
}
vector<ll> ruisekiwa(vector<ll>& a) {
vector<ll>rui(a.size() + 1);
rui[0] = 0;
for (ll i = 0; i < a.size(); ++i) {
rui[i + 1] = rui[i] + a[i];
}
return rui;
}
vector<bool>eratos(ll kaz) {
vector<bool>isSosu(kaz+1,true);
isSosu[0] = false; isSosu[1] = false;
for (ll i = 2; i * i <= kaz; ++i) {
if (isSosu[i]) {
for (ll j = i + i; j <= kaz; j += i) {
isSosu[j] = false;
}
}
}
return isSosu;
}
string yn(bool maj) {
return (maj ? "YES" : "NO");
}
template <typename T, typename U>typename common_type<T, U>::type max(const T& a, const U& b) { return a > b ? static_cast<typename common_type<T, U>::type>(a) : static_cast<typename common_type<T, U>::type>(b); }
template <typename T, typename U>typename common_type<T, U>::type min(const T& a, const U& b) { return a < b ? static_cast<typename common_type<T, U>::type>(a) : static_cast<typename common_type<T, U>::type>(b); }
//cpp_int
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define cy cout << "Yes" << endl, exit(0)
#define cn cout << "No" << endl, exit(0)
//reduce(all(a)) 全ての和
ll op(ll a, ll b) { return max(a, b); }
ll e() { return -INFi; }
struct phash {
inline size_t operator()(const pair<ll, ll>& p) const {
const auto h1 = hash<ll>()(p.first);
const auto h2 = hash<ll>()(p.second);
// p.firstのハッシュ値と、p.secondのハッシュ値をビットシフトさせたハッシュ値をXORする（シフトしないと(a,a)と(b,b), (a,b)と(b,a)などを渡した時同じハッシュ値が返ってしまうため）
return h1 ^ (h2 << 1);
}
};
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); cout << fixed << setprecision(20);
ll a, b, c, d, e;
cin >> a >> b >> c >> d >> e;
if (b == d)cy; cn;
}