题意：

有一个长度为\(n(1 \leq n \leq 10^5)\)的序列\(a\)
询问区间\([l,r]\)中不超过\(h\)的数字的个数

分析：

离线查询，将所有的\(a_i\)和\(h_i\)离散化，然后直接查询就可以了。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;const int maxn = 100000 + 10;struct Node
{int lch, rch, sum;
};int sz, root[maxn];
Node T[maxn << 5];int n, m;
int a[maxn], b[maxn * 2], tot;
int l[maxn], r[maxn], h[maxn];int newnode() {++sz;T[sz].lch = T[sz].rch = T[sz].sum = 0;return sz;
}int insert(int pre, int L, int R, int pos) {int rt = newnode();T[rt].lch = T[pre].lch;T[rt].rch = T[pre].rch;T[rt].sum = T[pre].sum + 1;if(L < R) {int M = (L + R) / 2;if(pos <= M) T[rt].lch = insert(T[pre].lch, L, M, pos);else T[rt].rch = insert(T[pre].rch, M+1, R, pos);}return rt;
}int query(int s, int t, int L, int R, int v) {if(v >= R) return T[t].sum - T[s].sum;int M = (L + R) / 2;if(v <= M) return query(T[s].lch, T[t].lch, L, M, v);return T[T[t].lch].sum - T[T[s].lch].sum + query(T[s].rch, T[t].rch, M+1, R, v);
}int main()
{int _; scanf("%d", &_);for(int kase = 1; kase <= _; kase++) {printf("Case %d:\n", kase);scanf("%d%d", &n, &m);for(int i = 1; i <= n; i++) {scanf("%d", a + i);b[i - 1] = a[i];}tot = n;for(int i = 1; i <= m; i++) {scanf("%d%d%d", l + i, r + i, h + i);b[tot++] = h[i];}sort(b, b + tot);tot = unique(b, b + tot) - b;for(int i = 1; i <= n; i++)a[i] = lower_bound(b, b + tot, a[i]) - b + 1;for(int i = 1; i <= m; i++)h[i] = lower_bound(b, b + tot, h[i]) - b + 1;sz = 0;for(int i = 1; i <= n; i++)root[i] = insert(root[i - 1], 1, tot, a[i]);for(int i = 1; i <= m; i++) {r[i]++;printf("%d\n", query(root[l[i]], root[r[i]], 1, tot, h[i]));}}return 0;
}