結果
問題 | No.989 N×Mマス計算(K以上) |
ユーザー | tran0826 |
提出日時 | 2020-07-01 09:09:41 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 27 ms / 2,000 ms |
コード長 | 10,204 bytes |
コンパイル時間 | 2,059 ms |
コンパイル使用メモリ | 149,984 KB |
実行使用メモリ | 6,944 KB |
最終ジャッジ日時 | 2024-09-13 20:50:02 |
合計ジャッジ時間 | 3,473 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge4 |
(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
6,812 KB |
testcase_01 | AC | 2 ms
6,812 KB |
testcase_02 | AC | 2 ms
6,820 KB |
testcase_03 | AC | 2 ms
6,944 KB |
testcase_04 | AC | 2 ms
6,940 KB |
testcase_05 | AC | 2 ms
6,940 KB |
testcase_06 | AC | 2 ms
6,940 KB |
testcase_07 | AC | 2 ms
6,940 KB |
testcase_08 | AC | 2 ms
6,944 KB |
testcase_09 | AC | 2 ms
6,940 KB |
testcase_10 | AC | 19 ms
6,944 KB |
testcase_11 | AC | 18 ms
6,944 KB |
testcase_12 | AC | 20 ms
6,940 KB |
testcase_13 | AC | 12 ms
6,944 KB |
testcase_14 | AC | 27 ms
6,940 KB |
testcase_15 | AC | 9 ms
6,940 KB |
testcase_16 | AC | 14 ms
6,944 KB |
testcase_17 | AC | 12 ms
6,940 KB |
testcase_18 | AC | 10 ms
6,944 KB |
testcase_19 | AC | 17 ms
6,944 KB |
ソースコード
#include<iostream> #include<vector> #include<set> #include<queue> #include<map> #include<algorithm> #include<cstring> #include<string> #include<cassert> #include<cmath> #include<climits> #include<iomanip> #include<stack> #include<unordered_map> #include<bitset> #include<limits> #include<complex> #include<array> #include<numeric> #include<functional> using namespace std; #define ll long long #define ull unsigned long long #define rrep(i,m,n) for(ll (i)=(ll)(m);(i)>=(ll)(n);(i)--) #define rep(i,m,n) for(ll (i)=(ll)(m);i<(ll)(n);i++) #define REP(i,n) rep(i,0,n) #define FOR(i,c) for(decltype((c).begin())i=(c).begin();i!=(c).end();++i) #define all(hoge) (hoge).begin(),(hoge).end() typedef pair<ll, ll> P; constexpr long double m_pi = 3.1415926535897932L; constexpr ll MOD = 1000000007; constexpr ll INF = 1LL << 62; constexpr long double EPS = 1e-10; template<typename T> using vector2 = vector<vector<T>>; template<typename T> using vector3 = vector<vector2<T>>; typedef vector<ll> Array; typedef vector<Array> Matrix; string operator*(const string& s, int k) { if (k == 0) return ""; string p = (s + s) * (k / 2); if (k % 2 == 1) p += s; return p; } template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; } template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; } struct Edge {//グラフ int to, rev; ll cap; Edge(int _to, ll _cap, int _rev) { to = _to; cap = _cap; rev = _rev; } }; typedef vector<Edge> Edges; typedef vector<Edges> Graph; void add_edge(Graph& G, int from, int to, ll cap, bool revFlag, ll revCap) {//最大フロー求める Ford-fulkerson G[from].push_back(Edge(to, cap, (ll)G[to].size() + (from == to))); if (revFlag)G[to].push_back(Edge(from, revCap, (ll)G[from].size() - 1));//最小カットの場合逆辺は0にする } ll max_flow_dfs(Graph& G, ll v, ll t, ll f, vector<bool>& used) { if (v == t) return f; used[v] = true; for (int i = 0; i < G[v].size(); ++i) { Edge& e = G[v][i]; if (!used[e.to] && e.cap > 0) { ll d = max_flow_dfs(G, e.to, t, min(f, e.cap), used); if (d > 0) { e.cap -= d; G[e.to][e.rev].cap += d; return d; } } } return 0; } //二分グラフの最大マッチングを求めたりも出来る また二部グラフの最大独立集合は頂点数-最大マッチングのサイズ ll max_flow(Graph& G, ll s, ll t)//O(V(V+E)) { ll flow = 0; for (;;) { vector<bool> used(G.size()); REP(i, used.size())used[i] = false; ll f = max_flow_dfs(G, s, t, INF, used); if (f == 0) { return flow; } flow += f; } } void BellmanFord(Graph& G, ll s, Array& d, Array& negative) {//O(|E||V|) d.resize(G.size()); negative.resize(G.size()); REP(i, d.size())d[i] = INF; REP(i, d.size())negative[i] = false; d[s] = 0; REP(k, G.size() - 1) { REP(i, G.size()) { REP(j, G[i].size()) { if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) { d[G[i][j].to] = d[i] + G[i][j].cap; } } } } REP(k, G.size() - 1) { REP(i, G.size()) { REP(j, G[i].size()) { if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) { d[G[i][j].to] = d[i] + G[i][j].cap; negative[G[i][j].to] = true; } if (negative[i] == true)negative[G[i][j].to] = true; } } } } void Dijkstra(Graph& G, ll s, Array& d) {//O(|E|log|V|) d.resize(G.size()); REP(i, d.size())d[i] = INF; d[s] = 0; priority_queue<P, vector<P>, greater<P>> q; q.push(make_pair(0, s)); while (!q.empty()) { P a = q.top(); q.pop(); if (d[a.second] < a.first)continue; REP(i, G[a.second].size()) { Edge e = G[a.second][i]; if (d[e.to] > d[a.second] + e.cap) { d[e.to] = d[a.second] + e.cap; q.push(make_pair(d[e.to], e.to)); } } } } void WarshallFloyd(Graph& G, Matrix& d) {//O(V^3) d.resize(G.size()); REP(i, d.size())d[i].resize(G.size()); REP(i, d.size()) { REP(j, d[i].size()) { d[i][j] = ((i != j) ? INF : 0); } } REP(i, G.size()) { REP(j, G[i].size()) { chmin(d[i][G[i][j].to], G[i][j].cap); } } REP(i, G.size()) { REP(j, G.size()) { REP(k, G.size()) { chmin(d[j][k], d[j][i] + d[i][k]); } } } } bool tsort(Graph& graph, Array& order) {//トポロジカルソートO(E+V) int n = graph.size(), k = 0; Array in(n); for (auto& es : graph) for (auto& e : es)in[e.to]++; priority_queue<ll, Array, greater<ll>> que; REP(i, n) if (in[i] == 0)que.push(i); while (que.size()) { int v = que.top(); que.pop(); order.push_back(v); for (auto& e : graph[v]) if (--in[e.to] == 0)que.push(e.to); } if (order.size() != n)return false; else return true; } class Lca { public: const int n = 0; const int log2_n = 0; std::vector<std::vector<int>> parent; std::vector<int> depth; Lca() {} Lca(const Graph& g, int root) : n(g.size()), log2_n(log2(n) + 1), parent(log2_n, std::vector<int>(n)), depth(n) { dfs(g, root, -1, 0); for (int k = 0; k + 1 < log2_n; k++) { for (int v = 0; v < (int)g.size(); v++) { if (parent[k][v] < 0) parent[k + 1][v] = -1; else parent[k + 1][v] = parent[k][parent[k][v]]; } } } void dfs(const Graph& g, int v, int p, int d) { parent[0][v] = p; depth[v] = d; for (auto& e : g[v]) { if (e.to != p) dfs(g, e.to, v, d + 1); } } int get(int u, int v) { if (depth[u] > depth[v]) std::swap(u, v); for (int k = 0; k < log2_n; k++) { if ((depth[v] - depth[u]) >> k & 1) { v = parent[k][v]; } } if (u == v) return u; for (int k = log2_n - 1; k >= 0; k--) { if (parent[k][u] != parent[k][v]) { u = parent[k][u]; v = parent[k][v]; } } return parent[0][u]; } }; class UnionFind { vector<int> data; ll num; public: UnionFind(int size) : data(size, -1), num(size) { } bool unite(int x, int y) {//xとyの集合を統合する x = root(x); y = root(y); if (x != y) { if (data[y] < data[x]) swap(x, y); data[x] += data[y]; data[y] = x; } num -= (x != y); return x != y; } bool findSet(int x, int y) {//xとyが同じ集合か返す return root(x) == root(y); } int root(int x) {//xのルートを返す return data[x] < 0 ? x : data[x] = root(data[x]); } ll size(int x) {//xの集合のサイズを返す return -data[root(x)]; } ll numSet() {//集合の数を返す return num; } }; template<typename T, typename F> class SegmentTree { private: T identity; F merge; ll n; vector<T> dat; public: SegmentTree(F f, T id,vector<T> v) :merge(f), identity(id) { int _n = v.size(); n = 1; while (n < _n)n *= 2; dat.resize(2 * n - 1, identity); REP(i, _n)dat[n + i - 1] = v[i]; for (int i = n - 2; i >= 0; i--)dat[i] = merge(dat[i * 2 + 1], dat[i * 2 + 2]); } SegmentTree(F f, T id, int _n) :merge(f), identity(id) { n = 1; while (n < _n)n *= 2; dat.resize(2 * n - 1, identity); } void set_val(int i, T x) { i += n - 1; dat[i] = x; while (i > 0) { i = (i - 1) / 2; dat[i] = merge(dat[i * 2 + 1], dat[i * 2 + 2]); } } T query(int l, int r) { T left = identity, right = identity; l += n - 1; r += n - 1; while (l < r) { if ((l & 1) == 0)left = merge(left, dat[l]); if ((r & 1) == 0)right = merge(dat[r - 1], right); l = l / 2; r = (r - 1) / 2; } return merge(left, right); } }; template< typename T > class FenwickTree { vector< T > data; int n; int p; public: FenwickTree(int n) :n(n) { data.resize(n + 1LL, 0); p = 1; while (p < data.size())p *= 2; } T sum(int k) { T ret = 0; for (; k > 0; k -= k & -k) ret += data[k]; return (ret); } T sum(int a, int b) { return sum(b) - sum(a); }//[a,b) void add(int k, T x) { for (++k; k <= n; k += k & -k) data[k] += x; } int lower_bound(ll w) { if (w <= 0)return -1; int x = 0; for (int k = p / 2; k > 0; k /= 2) { if (x + k <= n && data[x + k] < w)w -= data[x + k], x += k; } return x; } }; //約数求める //約数 void divisor(ll n, vector<ll>& ret) { for (ll i = 1; i * i <= n; i++) { if (n % i == 0) { ret.push_back(i); if (i * i != n) ret.push_back(n / i); } } sort(ret.begin(), ret.end()); } void prime_factorization(ll n, vector<P>& ret) { for (ll i = 2; i * i <= n; i++) { if (n % i == 0) { ret.push_back({ i,0 }); while (n % i == 0) { n /= i; ret[ret.size() - 1].second++; } } } if (n != 1)ret.push_back({ n,1 }); } inline ll mod_pow(ll x, ll n, ll mod) { ll res = 1; while (n > 0) { if (n & 1) res = res * x % mod; x = x * x % mod; n >>= 1; } return res; } inline ll mod_inv(ll x, ll mod) { return mod_pow(x, mod - 2, mod); } class Combination { public: Array fact; Array fact_inv; ll mod; //if n >= mod use lucas ll nCr(ll n, ll r) { if (n < r)return 0; if (n < mod)return ((fact[n] * fact_inv[r] % mod) * fact_inv[n - r]) % mod; ll ret = 1; while (n || r) { ll _n = n % mod, _r = r % mod; n /= mod; r /= mod; (ret *= nCr(_n, _r)) %= mod; } return ret; } ll nPr(ll n, ll r) { return (fact[n] * fact_inv[n - r]) % mod; } ll nHr(ll n, ll r) { return nCr(r + n - 1, r); } Combination(ll _n, ll _mod) { mod = _mod; ll n = min(_n + 1, mod); fact.resize(n); fact[0] = 1; REP(i, n - 1) { fact[i + 1] = (fact[i] * (i + 1LL)) % mod; } fact_inv.resize(n); fact_inv[n - 1] = mod_inv(fact[n - 1], mod); for (int i = n - 1; i > 0; i--) { fact_inv[i - 1] = fact_inv[i] * i % mod; } } }; ll popcount(ll x) { x = (x & 0x5555555555555555) + (x >> 1 & 0x5555555555555555); x = (x & 0x3333333333333333) + (x >> 2 & 0x3333333333333333); x = (x & 0x0F0F0F0F0F0F0F0F) + (x >> 4 & 0x0F0F0F0F0F0F0F0F); x = (x & 0x00FF00FF00FF00FF) + (x >> 8 & 0x00FF00FF00FF00FF); x = (x & 0x0000FFFF0000FFFF) + (x >> 16 & 0x0000FFFF0000FFFF); x = (x & 0x00000000FFFFFFFF) + (x >> 32 & 0x00000000FFFFFFFF); return x; } int main() { ios::sync_with_stdio(false); std::cin.tie(0); std::cout.tie(0); ll n, m, k; cin >> n >> m >> k; char op; cin >> op; Array b(m), a(n); REP(i, m)cin >> b[i]; REP(i, n)cin >> a[i]; sort(all(b),greater<ll>()); ll ans = 0; REP(i, n) { auto itr = partition_point(all(b), [&](ll b) {return (op == '*' ? a[i] * b : a[i] + b) >= k; }); ans += itr - b.begin(); } cout << ans << "\n"; return 0; }