只需要把询问按x升序排序，在查询的过程中不断让树状数组把<=x元素的下标处+1即可。（为此，把序列按val排序）

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;

pair<int, int> a[N];
#define val first
#define pos second

struct query {
    int l, r, x, pos, ans;
} q[N];

int n, m;

struct BIT {
    int t[N];
    BIT() { memset(t, 0, sizeof(t)); }
    void add(int i, int x) {
        for (; i <= n; i += i & -i)
            t[i] += x;
    }
    int sum(int i) {
        int res = 0;
        for (; i; i -= i & -i)
            res += t[i];
        return res;
    }
};

int main() {
    ios::sync_with_stdio(false); 
    cin.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        cin >> a[i].val;
        a[i].pos = i;
    }
    for (int i = 1; i <= m; i++) {
        cin >> q[i].l >> q[i].r >> q[i].x;
        q[i].pos = i; q[i].ans = 0;
    }
    sort(a + 1, a + 1 + n);
    sort(q + 1, q + 1 + m, [&](query a, query b) { return a.x < b.x; });
    
    BIT T; int p = 0;
    for (int i = 1; i <= m; i++) {
        query& Q = q[i];
        while (a[p + 1].val <= Q.x && p + 1 <= n) {
            p++;
            T.add(a[p].pos, 1);
        }
        Q.ans = T.sum(Q.r) - T.sum(Q.l - 1);
    }
    sort(q + 1, q + 1 + m, [&](query a, query b) { return a.pos < b.pos; });
    for (int i = 1; i <= m; i++)
        cout << q[i].ans << "\n";
    return 0;
}