結果

問題 No.694 square1001 and Permutation 3
ユーザー first_vilfirst_vil
提出日時 2020-06-04 22:30:33
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,170 ms / 3,000 ms
コード長 5,655 bytes
コンパイル時間 2,524 ms
コンパイル使用メモリ 211,900 KB
最終ジャッジ日時 2025-01-10 21:26:57
ジャッジサーバーID
（参考情報） 		judge1 / judge3
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
sample AC * 1
other AC * 13
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 # 

//四則演算 #pragma GCC target("avx")
//並列計算 #pragma GCC optimize("O3")
//条件分岐を減らす #pragma GCC optimize("unroll-loops")
//浮動小数点演算 #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using VI = vector<int>;
using VL = vector<ll>;
using VS = vector<string>;
template<class T> using PQ = priority_queue<T, vector<T>, greater<T>>;
#define FOR(i,a,n) for(int i=(a);i<(n);++i)
#define eFOR(i,a,n) for(int i=(a);i<=(n);++i)
#define rFOR(i,a,n) for(int i=(n)-1;i>=(a);--i)
#define erFOR(i,a,n) for(int i=(n);i>=(a);--i)
#define each(i, a) for(auto &i : a)
#define SORT(i) sort(i.begin(),i.end())
#define rSORT(i) sort(i.rbegin(),i.rend())
#define fSORT(i,a) sort(i.begin(),i.end(),a)
#define all(i) i.begin(),i.end()
#define out(y,x) ((y)<0||h<=(y)||(x)<0||w<=(x))
#define line cout << "-----------------------------\n" 
#define ENDL(i,n) ((i) == (n) - 1 ? "\n" : " ")
#define stop system("pause")
constexpr ll INF = 1000000000;
constexpr ll LLINF = 1LL << 60;
constexpr ll mod = 1000000007;
constexpr ll MOD = 998244353;
constexpr ld eps = 1e-10;
constexpr ld pi = 3.1415926535897932;
template<class T>inline bool chmax(T& a, const T& b) { if (a < b) { a = b; return true; }return false; }
template<class T>inline bool chmin(T& a, const T& b) { if (a > b) { a = b; return true; }return false; }
inline void init() { cin.tie(nullptr); cout.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); }
template<class T>inline istream& operator>>(istream& is, vector<T>& v) { for (auto& a : v)is >> a; return is; }
template<class T>inline istream& operator>>(istream& is, deque<T>& v) { for (auto& a : v)is >> a; return is; }
template<class T, class U>inline istream& operator>>(istream& is, pair<T, U>& p) { is >> p.first >> p.second; return is; }
template<class T>inline vector<T> vec(size_t a) { return vector<T>(a); }
template<class T>inline vector<T> defvec(T def, size_t a) { return vector<T>(a, def); }
template<class T, class... Ts>inline auto vec(size_t a, Ts... ts) { return vector<decltype(vec<T>(ts...))>(a, vec<T>(ts...)); }
template<class T, class... Ts>inline auto defvec(T def, size_t a, Ts... ts) { return vector<decltype(defvec<T>(def, ts...))>(a, defvec<T>(def, ts...)); }
template<class T>inline void print(const T& a) { cout << a << "\n"; }
template<class T, class... Ts>inline void print(const T& a, const Ts&... ts) { cout << a << " "; print(ts...); }
template<class T>inline void print(const vector<T>& v) { for (int i = 0; i < v.size(); ++i)cout << v[i] << (i == v.size() - 1 ? "\n" : " "); }
template<class T>inline void print(const vector<vector<T>>& v) { for (auto& a : v)print(a); }
inline string reversed(const string& s) { string t = s; reverse(all(t)); return t; }

template<class T> class segtree {
	using F1 = function<T(T, T)>;
	using F2 = function<void(T&, T)>;
	using F3 = function<bool(T, T)>;
	T unit;
	F1 f1;
	F2 f2;
	F3 f3;

	int n = 1, real_n;
	vector<T> dat;
	vector<pair<int, int>> range;
public:
	segtree() {}
	segtree(const int& _n, const F1& _f1, const F2& _f2, const F3& _f3, const T _unit = 0)
		: real_n(_n), f1(_f1), f2(_f2), f3(_f3), unit(_unit) {
		while (n < _n)n <<= 1;
		dat.resize(n << 1, unit), range.resize(n << 1);
		FOR(i, 0, n)range[n + i] = make_pair(i, i + 1);
		rFOR(i, 1, n)range[i] = make_pair(range[i << 1].first, range[i << 1 | 1].second);
	}
	segtree(const vector<T>& _dat, const F1& _f1, const F2& _f2, const F3& _f3, const T _unit = 0)
		: real_n(_dat.size()), f1(_f1), f2(_f2), f3(_f3), unit(_unit) {
		int _n = _dat.size();
		while (n < _n)n <<= 1;
		dat.resize(n << 1, unit), range.resize(n << 1);
		FOR(i, 0, _n)dat[i + n] = _dat[i];
		FOR(i, 0, n)range[n + i] = make_pair(i, i + 1);
		rFOR(i, 1, n) {
			dat[i] = f1(dat[i << 1], dat[i << 1 | 1]);
			range[i] = make_pair(range[i << 1].first, range[i << 1 | 1].second);
		}
	}
	void update(int i, T a) {
		i += n;
		f2(dat[i], a);
		while (1 < i) {
			i >>= 1;
			dat[i] = f1(dat[i << 1], dat[i << 1 | 1]);
		}
	}
	T value(int l, int r) {
		T ret = unit;
		for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
			if (l & 1)ret = f1(ret, dat[l++]);
			if (r & 1)ret = f1(ret, dat[--r]);
		}
		return ret;
	}
	int binary_search(T a) {
		if (!f3(dat[1], a))return real_n;
		int idx = 2;
		T cur = unit;
		for (; idx < n << 1; idx <<= 1) {
			if (!f3(f1(cur, dat[idx]), a)) {
				cur = f1(cur, dat[idx++]);
			}
		}
		return min((idx >> 1) - n, real_n);
	}
	int binary_search(int l, int r, T a) {
		T cur = unit;
		int ret = l;
		for (int idx = n + l; idx < n << 1 && ret < r;) {
			if (range[idx].second <= r && !f3(f1(cur, dat[idx]), a)) {
				cur = f1(cur, dat[idx]);
				ret = range[idx++].second;
				if (!(idx & 1))idx >>= 1;
			}
			else idx <<= 1;
		}
		return ret;
	}
	T operator[](int i) { return dat[i + n]; }
};

int main() {
    init();

    int n; cin >> n;
    VI a(n << 1);
    int s;
    {
        map<int, int> m;
        FOR(i, 0, n) {
            cin >> a[i];
            m[a[i]];
        }
        auto itr = m.begin();
        for (int i = 0; itr != m.end(); ++i, ++itr)itr->second = i;
        FOR(i, 0, n)a[i] = a[i + n] = m[a[i]];
        s = m.size();
    }

	segtree<ll> seg(s,
		[](ll x, ll y) {return x + y; },
		[](ll& x, ll y) {x += y; },
		[](ll x, ll y) {return true; });
	VL ans(n);
	FOR(i, 0, n) {
		ans[0] += seg.value(a[i] + 1, s);
		seg.update(a[i], 1);
	}
	FOR(i, 0, n - 1) {
		ans[i + 1] = ans[i];
		ans[i + 1] -= seg.value(0, a[i]);
		ans[i + 1] += seg.value(a[i] + 1, s);
	}
	FOR(i, 0, n)cout << ans[i] << "\n";

    return 0;
}
0