結果

問題 No.2374 ASKT Subsequences
ユーザー hibit_at
提出日時 2023-06-16 11:21:48
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 3,965 bytes
コンパイル時間 1,149 ms
コンパイル使用メモリ 110,748 KB
最終ジャッジ日時 2025-02-14 03:41:23
ジャッジサーバーID
（参考情報） 		judge2 / judge4
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 * 3
other AC * 2 WA * 26
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 <algorithm>
#include <iostream>
#include <iomanip>
#include <limits.h>
#include <map>
#include <math.h>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <string>
#include <utility>
#include <vector>
#include <stack>
#include <complex>

using namespace std;
#define rep(i, n) for (int i = 0; i < n; i++)
#define rep1(i, n) for (int i = 1; i < n + 1; i++)
#define rev(i, n) for (int i = n - 1; i >= 0; i--)
#define all(A) A.begin(), A.end()
#define itr(A, l, r) A.begin() + l, A.begin() + r
#define debug(var) cout << #var << " = " << var << endl;
typedef long long ll;

template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p)
{
    os << "(" << p.first << "," << p.second << ")";
    return os;
}

template <typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p)
{
    is >> p.first >> p.second;
    return is;
}

template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v)
{
    for (int i = 0; i < (int)v.size(); i++)
    {
        os << v[i] << (i + 1 != (int)v.size() ? " " : "");
    }
    return os;
}

template <typename T>
ostream &operator<<(ostream &os, const vector<vector<T>> &v)
{
    for (int i = 0; i < (int)v.size(); i++)
    {
        os << v[i] << endl;
    }
    return os;
}

template <typename T>
ostream &operator<<(ostream &os, const vector<vector<vector<T>>> &v)
{
    for (int i = 0; i < (int)v.size(); i++)
    {
        os << "i = " << i << endl;
        os << v[i];
    }
    return os;
}

template <typename T>
istream &operator>>(istream &is, vector<T> &v)
{
    for (T &in : v)
        is >> in;
    return is;
}

template <typename T, typename S>
ostream &operator<<(ostream &os, map<T, S> &mp)
{
    for (auto &[key, val] : mp)
    {
        os << key << ":" << val << " ";
    }
    cout << endl;
    return os;
}

template <typename T>
ostream &operator<<(ostream &os, set<T> st)
{
    auto itr = st.begin();
    for (int i = 0; i < (int)st.size(); i++)
    {
        os << *itr << (i + 1 != (int)st.size() ? " " : "");
        itr++;
    }
    return os;
}

template <typename T>
ostream &operator<<(ostream &os, multiset<T> st)
{
    auto itr = st.begin();
    for (int i = 0; i < (int)st.size(); i++)
    {
        os << *itr << (i + 1 != (int)st.size() ? " " : "");
        itr++;
    }
    return os;
}

template <typename T>
ostream &operator<<(ostream &os, queue<T> q)
{
    while (q.size())
    {
        os << q.front() << " ";
        q.pop();
    }
    return os;
}

template <typename T>
ostream &operator<<(ostream &os, deque<T> q)
{
    while (q.size())
    {
        os << q.front() << " ";
        q.pop_front();
    }
    return os;
}

template <typename T>
ostream &operator<<(ostream &os, stack<T> st)
{
    while (st.size())
    {
        os << st.top() << " ";
        st.pop();
    }
    return os;
}

template <typename T>
ostream &operator<<(ostream &os, priority_queue<T> pq)
{
    while (pq.size())
    {
        os << pq.top() << " ";
        pq.pop();
    }
    return os;
}

template <typename T>
ostream &operator<<(ostream &os, priority_queue<T, vector<T>, greater<T>> mpq)
{
    while (mpq.size())
    {
        os << mpq.top() << " ";
        mpq.pop();
    }
    return os;
}

int main()
{
    int n;
    cin >> n;
    vector<int> a(n);
    cin >> a;
    ll ans = 0;
    rep(d, 10)
    {
        // debug(d);
        vector<vector<int>> dp(n, vector<int>(4));
        dp[0][0] = 1;
        rep(i, n)
        {
            rep(k, i)
            {
                dp[i][1] += dp[k][0] * (a[k] == a[i] - d - 10);
            }
            rep(k, i)
            {
                dp[i][2] += dp[k][1] * (a[k] == a[i] + d);
            }
            rep(k, i)
            {
                dp[i][3] += dp[k][2] * (a[k] == a[i] - d - 1);
            }
        }
        // cout << dp << endl;
        rep(i, n)
        {
            ans += dp[i][3];
        }
    }
    cout << ans << endl;
}
0