結果

問題 No.924 紲星
ユーザー 👑 hitonanodehitonanode
提出日時 2019-11-09 18:17:17
言語 C++14
(gcc 13.2.0 + boost 1.83.0)
結果
TLE  
(最新)
AC  
(最初)
実行時間 -
コード長 5,370 bytes
コンパイル時間 2,495 ms
コンパイル使用メモリ 185,000 KB
実行使用メモリ 31,164 KB
最終ジャッジ日時 2023-10-13 07:26:39
合計ジャッジ時間 35,012 ms
ジャッジサーバーID
（参考情報） 		judge12 / judge15
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 1 ms
4,348 KB
testcase_01 AC 1 ms
4,348 KB
testcase_02 AC 1 ms
4,348 KB
testcase_03 AC 7 ms
4,352 KB
testcase_04 AC 4 ms
4,352 KB
testcase_05 AC 9 ms
4,352 KB
testcase_06 AC 8 ms
4,352 KB
testcase_07 AC 4 ms
4,352 KB
testcase_08 TLE -
testcase_09 TLE -
testcase_10 TLE -
testcase_11 TLE -
testcase_12 TLE -
testcase_13 AC 1,776 ms
16,096 KB
testcase_14 AC 1,799 ms
18,304 KB
testcase_15 AC 1,585 ms
14,156 KB
testcase_16 AC 1,928 ms
18,484 KB
testcase_17 AC 2,734 ms
21,760 KB
testcase_18 AC 2 ms
4,352 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>
using namespace std;
#pragma GCC optimize("O3", "unroll-loops")
#pragma GCC target("avx")
using lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((lint)(x).size())
#define POW2(n) (1LL << (n))
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template<typename T> void ndarray(vector<T> &vec, int len) { vec.resize(len); }
template<typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) ndarray(v, args...); }
template<typename T> bool mmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template<typename T> bool mmin(T &m, const T q) { if (m > q) {m = q; return true;} else return false; }
template<typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template<typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
template<typename T> istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; }
///// This part below is only for debug, not used /////
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec){ os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const deque<T> &vec){ os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa){ os << "(" << pa.first << "," << pa.second << ")"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;
///// END /////

vector<lint> A, Aacc;
// 1-indexed BIT (i : [1, len])
int N, Q;
struct BIT
{
    using T = lint;
    int len;
    vector<T> val;
    BIT(int num) : len(num), val(num + 1) {}
    BIT() : BIT(0) {}
    void reset() { fill(val.begin(), val.end(), 0); }
    void add(int pos, T v) { while (pos > 0 and pos <= len) val[pos] += v, pos += pos & -pos; }
    T sum(int pos) const // (0, pos]
    {
        T res = 0;
        while (pos > 0) res += val[pos], pos -= pos & -pos;
        return res;
    }
};
template<typename T> ostream &operator<<(ostream &os, const BIT &b){ T prv = 0; os << "["; REP(i, b.len) { T nxt = b.sum(i + 1); os << nxt - prv << ","; prv = nxt; } os << "]"; return os; }

BIT cou, tot;
vector<int> L, R;
vector<plint> solve(const vector<lint> &c)
{
    cou = BIT(N + 1);
    tot = BIT(N + 1);
    vector<plint> ret(Q);
    priority_queue<plint, vector<plint>, greater<plint>> pq, as;
    REP(i, Q) pq.emplace(c[i], i);
    REP(i, N) as.emplace(A[i], i);
    while (!pq.empty())
    {
        lint q, i;
        tie(q, i) = pq.top();
        while (!as.empty() and as.top().first < q)
        {
            lint i, val;
            tie(val, i) = as.top();
            as.pop();
            cou.add(i + 1, 1);
            tot.add(i + 1, val);
        }
        pq.pop();
        ret[i].first = cou.sum(R[i]) - cou.sum(L[i]);
        ret[i].second = tot.sum(R[i]) - tot.sum(L[i]);
    }
    return ret;
}

int main()
{
    cin >> N >> Q;
    A.resize(N);
    cin >> A;
    Aacc.resize(N + 1);
    REP(i, N) Aacc[i + 1] = Aacc[i] + A[i];
    L.resize(Q);
    R.resize(Q);
    REP(i, Q) cin >> L[i] >> R[i];
    REP(i, Q) L[i]--;
    vector<lint> lo(Q, -1e9 - 10), hi(Q, 1e9 + 10);
    REP(_, 31)
    {
        vector<lint> c(Q);
        REP(i, Q) c[i] = (lo[i] + hi[i]) / 2;
        vector<plint> ret = solve(c);
        REP(i, Q)
        {
            if (ret[i].first <= (R[i] - L[i]) / 2) lo[i] = c[i];
            else hi[i] = c[i];
        }
    }
    vector<plint> ret = solve(lo);
    REP(i, Q)
    {
        lint m = lo[i];
        lint n = R[i] - L[i];
        lint T = Aacc[R[i]] - Aacc[L[i]];
        printf("%lld\n", (T - ret[i].second - (n - ret[i].first) * m) + ret[i].first * m - ret[i].second);
    }
}
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