結果

問題 No.2956 Substitute with Average
ユーザー 👑 binapbinap
提出日時 2024-10-20 13:48:46
言語 C++23(gcc13)
(gcc 13.2.0 + boost 1.83.0)
結果
AC  
実行時間 83 ms / 3,000 ms
コード長 2,993 bytes
コンパイル時間 7,871 ms
コンパイル使用メモリ 335,440 KB
実行使用メモリ 51,444 KB
最終ジャッジ日時 2024-10-25 19:44:53
合計ジャッジ時間 8,214 ms
ジャッジサーバーID
(参考情報)
judge5 / judge4
このコードへのチャレンジ
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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
6,812 KB
testcase_01 AC 2 ms
6,812 KB
testcase_02 AC 2 ms
6,816 KB
testcase_03 AC 2 ms
6,820 KB
testcase_04 AC 2 ms
6,816 KB
testcase_05 AC 2 ms
6,820 KB
testcase_06 AC 2 ms
6,820 KB
testcase_07 AC 2 ms
6,820 KB
testcase_08 AC 2 ms
6,820 KB
testcase_09 AC 2 ms
6,820 KB
testcase_10 AC 69 ms
51,200 KB
testcase_11 AC 70 ms
50,908 KB
testcase_12 AC 69 ms
51,052 KB
testcase_13 AC 69 ms
51,376 KB
testcase_14 AC 71 ms
50,976 KB
testcase_15 AC 69 ms
51,208 KB
testcase_16 AC 68 ms
51,248 KB
testcase_17 AC 68 ms
50,948 KB
testcase_18 AC 73 ms
51,440 KB
testcase_19 AC 69 ms
51,444 KB
testcase_20 AC 72 ms
51,440 KB
testcase_21 AC 69 ms
51,440 KB
testcase_22 AC 83 ms
51,316 KB
testcase_23 AC 78 ms
51,440 KB
testcase_24 AC 69 ms
51,320 KB
testcase_25 AC 75 ms
51,436 KB
testcase_26 AC 67 ms
51,436 KB
testcase_27 AC 67 ms
51,316 KB
権限があれば一括ダウンロードができます

ソースコード

diff #

#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;
}
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