題意：有海象排隊，每個海象都有不滿意度，取決於離他前面的最遠的年齡比他小的海象和他直接站的海象數，給出佇列裡每個海象的年齡，輸出每個海象的不滿意度，如果沒有不滿意輸出-1。
題解：線段樹查詢最遠的比自己年齡小的海象的下標。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100050;
int n, a[N], tree[N << 2];

void pushup(int k) {
    tree[k] = min(tree[k * 2
 
], tree[k * 2 + 1]);
}

void build(int k, int left, int right) {
    if (left == right) {
        tree[k] = a[left];
        return;
    }
    int mid = (left + right) / 2;
    build(k * 2, left, mid);
    build(k * 2 + 1, mid + 1, right);
    pushup(k);
}

int query(int k, int left, int right, int l1, int
 
 r1, int x) {
    if (left == right) {
        if (tree[k] < x)
            return left;
        return -1;
    }
    int mid = (left + right) / 2;
    if (r1 > mid && tree[k * 2 + 1] < x)
        return query(k * 2 + 1, mid + 1, right, l1, r1, x);
    else if (l1 <= mid)
        return query(k * 2
 
, left, mid, l1, r1, x);
    return -1;
}

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
    build(1, 1, n);
    for (int i = 1; i < n; i++) {
        int temp = query(1, 1, n, i + 1, n, a[i]);
        if (temp == -1)
            printf("-1 ");
        else
            printf("%d ", temp - i - 1);
    }
    printf("-1\n");
    return 0;
}