結果

問題 No.2601 Very Poor
ユーザー hibit_at
提出日時 2024-01-13 00:32:03
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 94 ms / 2,000 ms
コード長 4,278 bytes
コンパイル時間 1,385 ms
コンパイル使用メモリ 107,188 KB
最終ジャッジ日時 2025-02-18 19:25:53
ジャッジサーバーID
（参考情報） 		judge2 / judge5
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 * 34
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)
{
    int n = v.size();
    int m = v[0].size();
    int p = v[0][0].size();
    rep(k, p)
    {
        os << "k = " << k << endl;
        rep(i, n)
        {
            rep(j, m)
            {
                os << v[i][j][k];
                if (j < m - 1)
                {
                    os << " ";
                }
                else
                {
                    os << endl;
                }
            }
        }
    }
    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()
{
    ll n, x;
    cin >> n >> x;
    vector<ll> a(n);
    cin >> a;
    rep(i, n)
    {
        a.push_back(a[i]);
    }
    vector<ll> s(2 * n + 1);
    rep(i, 2 * n)
    {
        s[i + 1] = s[i] + a[i];
    }
    // debug(s);
    if (x >= s[n])
    {
        cout << s[n] << endl;
        return 0;
    }
    ll ans = -1;
    rep(i, n)
    {
        auto itr = upper_bound(all(s), s[i] + x);
        if (itr == s.begin())
        {
            continue;
        }
        // debug(*itr);
        ll tmp = *prev(itr) - s[i];
        // debug(tmp);
        auto chmax = [](auto &a, auto b)
        { a = max(a, b); };
        chmax(ans, tmp);
    }
    cout << ans << endl;
}
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