結果
問題 | No.2956 Substitute with Average |
ユーザー | 👑 binap |
提出日時 | 2024-10-18 20:15:19 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 79 ms / 3,000 ms |
コード長 | 2,993 bytes |
コンパイル時間 | 4,482 ms |
コンパイル使用メモリ | 264,260 KB |
実行使用メモリ | 51,452 KB |
最終ジャッジ日時 | 2024-10-25 19:44:08 |
合計ジャッジ時間 | 7,195 ms |
ジャッジサーバーID (参考情報) |
judge3 / 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 | 2 ms
6,816 KB |
testcase_01 | AC | 2 ms
6,816 KB |
testcase_02 | AC | 2 ms
6,820 KB |
testcase_03 | AC | 2 ms
6,820 KB |
testcase_04 | AC | 2 ms
6,820 KB |
testcase_05 | AC | 2 ms
6,816 KB |
testcase_06 | AC | 2 ms
6,816 KB |
testcase_07 | AC | 2 ms
6,820 KB |
testcase_08 | AC | 2 ms
6,816 KB |
testcase_09 | AC | 2 ms
6,816 KB |
testcase_10 | AC | 69 ms
51,200 KB |
testcase_11 | AC | 69 ms
50,908 KB |
testcase_12 | AC | 69 ms
51,056 KB |
testcase_13 | AC | 69 ms
51,248 KB |
testcase_14 | AC | 68 ms
50,848 KB |
testcase_15 | AC | 69 ms
51,228 KB |
testcase_16 | AC | 70 ms
51,248 KB |
testcase_17 | AC | 68 ms
50,948 KB |
testcase_18 | AC | 73 ms
51,312 KB |
testcase_19 | AC | 68 ms
51,308 KB |
testcase_20 | AC | 72 ms
51,316 KB |
testcase_21 | AC | 69 ms
51,312 KB |
testcase_22 | AC | 79 ms
51,440 KB |
testcase_23 | AC | 77 ms
51,452 KB |
testcase_24 | AC | 67 ms
51,312 KB |
testcase_25 | AC | 68 ms
51,312 KB |
testcase_26 | AC | 67 ms
51,312 KB |
testcase_27 | AC | 70 ms
51,440 KB |
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);} int main(){ int n; cin >> n; vector<int> a(n); cin >> a; const int M = 30; const int R = n * M; vector<int> cnt_zero(2 * R + 1); vector<int> cnt_non_zero(2 * R + 1); long long ans = (long long)n * (n + 1) / 2; for(int x = 1; x <= M; x++){ vector<int> b(n); rep(i, n) b[i] = a[i] - x; vector<int> s(n + 1); rep(i, n) s[i + 1] = s[i] + b[i]; rep(i, n){ if(b[i] == 0){ cnt_zero[R + s[i]]++; ans -= cnt_zero[R + s[i + 1]] + cnt_non_zero[R + s[i + 1]]; }else{ cnt_non_zero[R + s[i]]++; ans -= cnt_zero[R + s[i + 1]]; } } rep(i, n){ if(b[i] == 0) cnt_zero[R + s[i]] = 0; else cnt_non_zero[R + s[i]] = 0; } } ans++; cout << ans << "\n"; return 0; }