結果
問題 | No.2210 equence Squence Seuence |
ユーザー | ktr216 |
提出日時 | 2023-02-10 21:59:15 |
言語 | C++14 (gcc 12.3.0 + boost 1.83.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 3,730 bytes |
コンパイル時間 | 1,668 ms |
コンパイル使用メモリ | 182,912 KB |
実行使用メモリ | 14,132 KB |
最終ジャッジ日時 | 2024-07-07 16:14:20 |
合計ジャッジ時間 | 4,404 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge1 |
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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 | 3 ms
9,536 KB |
testcase_01 | AC | 3 ms
9,668 KB |
testcase_02 | AC | 3 ms
9,540 KB |
testcase_03 | WA | - |
testcase_04 | WA | - |
testcase_05 | WA | - |
testcase_06 | AC | 13 ms
10,304 KB |
testcase_07 | AC | 30 ms
11,764 KB |
testcase_08 | WA | - |
testcase_09 | AC | 3 ms
9,540 KB |
testcase_10 | AC | 3 ms
9,544 KB |
testcase_11 | WA | - |
testcase_12 | AC | 3 ms
9,544 KB |
testcase_13 | AC | 39 ms
11,956 KB |
testcase_14 | WA | - |
testcase_15 | WA | - |
testcase_16 | AC | 37 ms
12,080 KB |
testcase_17 | AC | 38 ms
12,084 KB |
testcase_18 | WA | - |
testcase_19 | AC | 3 ms
9,796 KB |
testcase_20 | WA | - |
testcase_21 | AC | 3 ms
11,588 KB |
testcase_22 | AC | 2 ms
9,544 KB |
testcase_23 | RE | - |
testcase_24 | RE | - |
testcase_25 | RE | - |
testcase_26 | RE | - |
testcase_27 | RE | - |
ソースコード
#include <bits/stdc++.h> using namespace std; #define double long double using ll = long long; using VB = vector<bool>; using VVB = vector<VB>; using VVVB = vector<VVB>; using VC = vector<char>; using VVC = vector<VC>; using VI = vector<int>; using VVI = vector<VI>; using VVVI = vector<VVI>; using VVVVI = vector<VVVI>; using VL = vector<ll>; using VVL = vector<VL>; using VVVL = vector<VVL>; using VVVVL = vector<VVVL>; using VD = vector<double>; using VVD = vector<VD>; using VVVD = vector<VVD>; //using P = pair<int, int>; #define REP(i, n) for (int i = 0; i < (int)(n); i++) #define FOR(i, a, b) for (ll i = a; i < (ll)(b); i++) #define ALL(a) (a).begin(),(a).end() constexpr int INF = 1001001001; constexpr ll LINF = 1001001001001001001ll; constexpr int DX[] = {2, 1, -1, -2, -2, -1, 1, 2}; constexpr int DY[] = {1, 2, 2, 1, -1, -2, -2, -1}; template< typename T1, typename T2> inline bool chmax(T1 &a, T2 b) {return a < b && (a = b, true); } template< typename T1, typename T2> inline bool chmin(T1 &a, T2 b) {return a > b && (a = b, true); } const ll MOD = 998244353; const int MAX_N = 400010; int par[MAX_N]; int rnk[MAX_N]; int siz[MAX_N]; void init(int n) { REP(i,n) { par[i] = i; rnk[i] = 0; siz[i] = 1; } } int find(int x) { if (par[x] == x) { return x; } else { return par[x] = find(par[x]); } } void unite(int x, int y) { x = find(x); y = find(y); if (x == y) return; int s = siz[x] + siz[y]; if (rnk[x] < rnk[y]) { par[x] = y; } else { par[y] = x; if (rnk[x] == rnk[y]) rnk[x]++; } siz[find(x)] = s; } bool same(int x, int y) { return find(x) == find(y); } int size(int x) { return siz[find(x)]; } ll mod_pow(ll x, ll n, ll mod) { ll res = 1; x %= mod; while (n > 0) { if (n & 1) res = res * x % mod; x = x * x % mod; n >>= 1; } return res; } ll gcd(ll x, ll y) { if (y == 0) return x; return gcd(y, x % y); } typedef pair<double, int> P0; struct edge { int to; double cost; }; const int MAX_V = 200000; //const ll LINF = 1LL<<60; int V; vector<edge> G[MAX_V]; double d[MAX_V]; void dijkstra(ll s) { // greater<P>を指定することでfirstが小さい順に取り出せるようにする priority_queue<P0, vector<P0>, greater<P0> > que; fill(d, d + V, LINF); d[s] = 0; que.push(P0(0, s)); while (!que.empty()) { P0 p = que.top(); que.pop(); int v = p.second; if (d[v] < p.first) continue; REP(i,G[v].size()) { edge e = G[v][i]; if (d[e.to] > d[v] + e.cost) { d[e.to] = d[v] + e.cost; que.push(P0(d[e.to], e.to)); } } } } /* VL f(51, 1); ll C(ll n, ll k) { return f[n] * mod_pow(f[k], MOD - 2, MOD) % MOD * mod_pow(f[n - k], MOD - 2, MOD) % MOD; } */ int main() { ios::sync_with_stdio(false); std::cin.tie(nullptr); //REP(i, 50) f[i + 1] = f[i] * (i + 1) % MOD; int N, K; cin >> N >> K; VI A(N), B(0), C(0); REP(i, N) cin >> A[i]; deque<int> dq = {N - 1}; for (int i = N - 2; i >= 0; i--) { if (A[i] < A[i + 1]) { for (int j : C) dq.push_back(j); dq.push_back(i); C = {}; } else if (A[i] > A[i + 1]) { for (int j : C) dq.push_front(j); dq.push_front(i); C = {}; } else { C.push_back(i); } } int X = dq[K - 1]; REP(i, N) if (i != X) B.push_back(A[i]); REP(i, N - 1) { cout << B[i]; if (i < N - 2) cout << " "; } cout << endl; //for (int i : dq) cout << i << " "; cout << endl; }