結果
問題 |
No.3041 非対称じゃんけん
|
ユーザー |
👑 |
提出日時 | 2025-02-28 22:20:55 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 408 ms / 2,200 ms |
コード長 | 3,110 bytes |
コンパイル時間 | 4,152 ms |
コンパイル使用メモリ | 253,844 KB |
実行使用メモリ | 7,332 KB |
最終ジャッジ日時 | 2025-02-28 22:21:02 |
合計ジャッジ時間 | 6,845 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 1 |
other | AC * 30 |
ソースコード
#include<bits/stdc++.h> #include<atcoder/all> #define rep(i,n) for(int i=0;i<n;i++) using namespace std; using namespace atcoder; typedef long long ll; typedef vector<int> vi; typedef vector<long long> vl; typedef vector<vector<int>> vvi; typedef vector<vector<long long>> vvl; typedef long double ld; typedef pair<int, int> P; template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;} template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;} template <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;} template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;} template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;} template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;} template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;} template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;} template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;} template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;} template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template<typename T> void chmin(T& a, T b){a = min(a, b);} template<typename T> void chmax(T& a, T b){a = max(a, b);} const int INF = 1001001001; int main(){ int n, f; cin >> n >> f; vector<int> a(n); vector<int> b(n); vector<int> c(n); cin >> a; cin >> b; cin >> c; vector<vector<int>> idx(f + 1); rep(i, n){ vector<int> vec = {a[i], b[i], c[i]}; sort(vec.begin(), vec.end()); idx[vec[1] - vec[0]].push_back(i); idx[vec[2] - vec[0]].push_back(i); } vector<int> ans(n * f + 1,INF); vector<int> ans2(n); ans[0] = -1; for(int v = 0; v <= n * f; v++){ if(ans[v] == INF) continue; int threshold = ans[v]; for(int x = 1; x <= f; x++){ auto itr = upper_bound(idx[x].begin(), idx[x].end(), threshold); if(itr == idx[x].end()) continue; chmin(ans[v + x], *itr); } if(v == 0) ans2[0]++; else ans2[threshold]++; } for(int i = 1; i < n; i++) ans2[i] += ans2[i - 1]; rep(i, n) cout << ans2[i] << "\n"; return 0; }