题解 | #换个角度思考#

我最开始做的是线段树每个节点维护一个桶, 空间算错了, 直接爆了, 正确做法是主席树或者离线树状数组

将查询分为和, 分别查询累加贡献

#include <bits/stdc++.h>

#define x first
#define y second
#define all(x) x.begin(), x.end()
#define vec1(T, name, n, val) vector<T> name(n, val)
#define vec2(T, name, n, m, val) vector<vector<T>> name(n, vector<T>(m, val))
#define vec3(T, name, n, m, k, val) vector<vector<vector<T>>> name(n, vector<vector<T>>(m, vector<T>(k, val)))
#define vec4(T, name, n, m, k, p, val) vector<vector<vector<vector<T>>>> name((n), vector<vector<vector<T>>>((m), vector<vector<T>>((k), vector<T>((p), (val)))))

using namespace std;
using i128 = __int128;
using u128 = unsigned __int128;
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
using PII = pair<int, int>;
using PLL = pair<LL, LL>;
using PLD = pair<LD, LD>;

const int N = 1e5 + 10, MOD = 998244353;
const int INF = 1e9;
const LL LL_INF = 2e18;
const LD EPS = 1e-8;
const int dx4[] = {-1, 0, 1, 0}, dy4[] = {0, 1, 0, -1};
const int dx8[] = {-1, -1, -1, 0, 0, 1, 1, 1}, dy8[] = {-1, 0, 1, -1, 1, -1, 0, 1};
const int hx[] = {-2, -2, -1, -1, 1, 1, 2, 2}, hy[] = {-1, 1, -2, 2, -2, 2, -1, 1};

istream& operator>>(istream& is, i128& val) {
    string str;
    is >> str;
    val = 0;
    bool flag = false;
    if (str[0] == '-') flag = true, str = str.substr(1);
    for (char& c : str) val = val * 10 + c - '0';
    if (flag) val = -val;
    return is;
}

ostream& operator<<(ostream& os, i128 val) {
    if (val < 0) os << "-", val = -val;
    if (val > 9) os << val / 10;
    os << static_cast<char>(val % 10 + '0');
    return os;
}

bool cmp(LD a, LD b) {
    if (fabs(a - b) < EPS) return 1;
    return 0;
}

LL qpow(LL a, LL b) {
    LL ans = 1;
    a %= MOD;
    while (b) {
        if (b & 1) ans = ans * a % MOD;
        a = a * a % MOD;
        b >>= 1;
    }
    return ans;
}

struct Fenwick {
    int n;
    vector<LL> tr;

    Fenwick(int _n) : n(_n + 1), tr(_n + 1, 0) {}

    int lowbit(int x) {
        return x & -x;
    }

    void add(int u, LL x) {
        if (u <= 0) return;
        for (int i = u; i < n; i += lowbit(i)) tr[i] += x;
    }

    LL query(int u) {
        u = min(u, n - 1);
        LL ans = 0;
        for (int i = u; i; i -= lowbit(i)) ans += tr[i];
        return ans;
    }

    LL query(int a, int b) {
        if (a > b) return 0;
        return query(b) - query(a - 1);
    }

    LL kth(LL k) {
        int x = 0;
        for (int p = 1 << 20; p; p >>= 1) {
            if (x + p <= n && tr[x + p] < k) {
                k -= tr[x + p];
                x += p;
            }
        }
        return x + 1;
    }
};

void solve() {
    int n, m;
    cin >> n >> m;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; ++i) cin >> a[i];
    Fenwick tr(N);
    
    struct Q {
        int v;
        int k;
        int id;

        bool operator< (Q q) const {
            return v < q.v;
        }
    };
    vector<Q> qs;
    for (int i = 1; i <= m; ++i) {
        int l, r, k;
        cin >> l >> r >> k;
        if (k >= N) k = N - 1;
        qs.push_back({r, k, i});
        qs.push_back({l - 1, k, -i});
    }

    sort(all(qs));

    int idx = 0;
    vector<int> ans(m + 1);
    for (auto [pos, k, id] : qs) {
        while (idx < pos) idx++, tr.add(a[idx], 1);
        if (id > 0) ans[id] += tr.query(k);
        else ans[-id] -= tr.query(k);
    }

    for (int i = 1; i <= m; ++i) cout << ans[i] << '\n';

/**/ #ifdef LOCAL
    cout
        << flush;
/**/ #endif
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int T = 1;
    while (T--) solve();
    cout << fixed << setprecision(15);

    return 0;
}