結果

問題 No.2374 ASKT Subsequences
ユーザー ktr216
提出日時 2023-07-07 22:42:53
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 967 ms / 2,000 ms
コード長 4,351 bytes
コンパイル時間 2,054 ms
コンパイル使用メモリ 181,728 KB
実行使用メモリ 6,940 KB
最終ジャッジ日時 2024-07-21 18:58:13
合計ジャッジ時間 5,547 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 28
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 #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
#define double long double
using ll = long long;
using VB = vector<bool>;
using VVB = vector<VB>;
using VVVB = vector<VVB>;
using VC = vector<char>;
using VVC = vector<VC>;
using VI = vector<int>;
using VVI = vector<VI>;
using VVVI = vector<VVI>;
using VVVVI = vector<VVVI>;
using VL = vector<ll>;
using VVL = vector<VL>;
using VVVL = vector<VVL>;
using VVVVL = vector<VVVL>;
using VD = vector<double>;
using VVD = vector<VD>;
using VVVD = vector<VVD>;
//using P = pair<int, int>;
#define REP(i, n) for (ll i = 0; i < (int)(n); i++)
#define FOR(i, a, b) for (ll i = a; i < (ll)(b); i++)
#define ALL(a) (a).begin(),(a).end()
constexpr int INF = 1001001001;
constexpr ll LINF = 1001001001001001001ll;
constexpr int DX[] = {1, 0, -1, 0};
constexpr int DY[] = {0, 1, 0, -1};
template< typename T1, typename T2>
inline bool chmax(T1 &a, T2 b) {return a < b && (a = b, true); }
template< typename T1, typename T2>
inline bool chmin(T1 &a, T2 b) {return a > b && (a = b, true); }
const ll MOD = 998244353;
const int MAX_N = 40;
int par[MAX_N];
int rnk[MAX_N];
int siz[MAX_N];
void init(int n) {
    REP(i,n) {
        par[i] = i;
        rnk[i] = 0;
        siz[i] = 1;
    }
}
int find(int x) {
    if (par[x] == x) {
        return x;
    } else {
        return par[x] = find(par[x]);
    }
}
void unite(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y) return;
    int s = siz[x] + siz[y];
    if (rnk[x] < rnk[y]) {
        par[x] = y;
    } else {
        par[y] = x;
        if (rnk[x] == rnk[y]) rnk[x]++;
    }
    siz[find(x)] = s;
}
bool same(int x, int y) {
    return find(x) == find(y);
}
int size(int x) {
    return siz[find(x)];
}
ll mod_pow(ll x, ll n, ll mod) {
    ll res = 1;
    x %= mod;
    while (n > 0) {
        if (n & 1) res = res * x % mod;
        x = x * x % mod;
        n >>= 1;
    }
    return res;
}
ll gcd(ll x, ll y) {
    if (y == 0) return x;
    return gcd(y, x % y);
}
typedef pair<ll, int> P0;
struct edge { int to; ll cost; };
const int MAX_V = 502;
//const ll LINF = 1LL<<60;
int V;
vector<edge> G[MAX_V];
ll d[MAX_V];
void dijkstra(ll s) {
    priority_queue<P0, vector<P0>, greater<P0> > que;
    fill(d, d + V, LINF);
    d[s] = 0;
    que.push(P0(0, s));
    while (!que.empty()) {
        P0 p = que.top(); que.pop();
        int v = p.second;
        if (d[v] < p.first) continue;
        for (edge e : G[v]) {
            if (d[e.to] > d[v] + e.cost) {
                d[e.to] = d[v] + e.cost;
                que.push(P0(d[e.to], e.to));
            }
        }
    }
}
/*
double EPS = 1e-10;
double add(double a, double b) {
    if (abs(a + b) < EPS * (abs(a) + abs(b))) return 0;
    return a + b;
}
struct P {
    double x, y;
    P() {}
    P(double x, double y) : x(x), y(y) {
    }
    P operator + (P p) {
        return P(add(x, p.x), add(y, p.y));
    }
    P operator - (P p) {
        return P(add(x, -p.x), add(y, -p.y));
    }
    P operator * (double d) {
        return P(x * d, y * d);
    }
    double dot(P p) {
        return add(x * p.x, y * p.y);
    }
    double det(P p) {
        return add(x * p.y, -y * p.x);
    }
};
bool on_seg(P p1, P p2, P q) {
    return ()
}
P intersection(P p1, P p2, P q1, P q2) {
    return p1 + (p2 - p1) * ((q2 - q1).det(q1 - p1) / (q2 - q1).det(p2 - p1));
}
*/
/*
VL f(400010, 1);
ll C(ll n, ll k) {
    return f[n] * mod_pow(f[k], MOD - 2, MOD) % MOD * mod_pow(f[n - k], MOD - 2, MOD) % MOD;
}
*/
int main() {
    ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    //REP(i, 400009) f[i + 1] = f[i] * (i + 1) % MOD;
    int N;
    cin >> N;
    VI A(N);
    REP(i, N) cin >> A[i];
    VVI B(2001);
    REP(i, N) B[A[i]].push_back(i);
    ll ans = 0;
    FOR(a1, 1, 2001) FOR(a2, a1 + 11, 2001) {
        int a3 = a1 + 10, a4 = a2 + 1;
        if (a3 > 2000 || a4 > 2000) continue;
        REP(i, B[a1].size()) {
            int x2 = upper_bound(ALL(B[a2]), B[a1][i]) - B[a2].begin();
            FOR(j, x2, B[a2].size()) {
                int x3 = upper_bound(ALL(B[a3]), B[a2][j]) - B[a3].begin();
                FOR(k, x3, B[a1 + 10].size()) {
                    int x4 = upper_bound(ALL(B[a4]), B[a3][k]) - B[a4].begin();
                    ans += B[a4].size() - x4;
                }
            }
        }
    }
    cout << ans << endl;
}
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