CodeForces 91B Queue （线段树单点操作）

Description

There are n walruses standing in a queue in an airport. They are numbered starting from the queue‘s tail: the
1-st walrus stands at the end of the queue and the
n-th walrus stands at the beginning of the queue. The
i-th walrus has the age equal to a_i.

The i-th walrus becomes displeased if there‘s a younger walrus standing in front of him, that is, if exists such
j (i?<?j), that
a_i?>?a_j. The
displeasure of the
i-th walrus is equal to the number of walruses between him and the furthest walrus ahead of him, which is younger than the
i-th one. That is, the further that young walrus stands from him, the stronger the displeasure is.

The airport manager asked you to count for each of n walruses in the queue his displeasure.

Input

The first line contains an integer n (2?≤?n?≤?10⁵) — the number of walruses in the queue. The second line contains
integers a_i (1?≤?a_i?≤?10⁹).

Note that some walruses can have the same age but for the displeasure to emerge the walrus that is closer to the head of the queue needs to be
strictly younger than the other one.

Output

Print n numbers: if the
i-th walrus is pleased with everything, print "-1" (without the quotes). Otherwise, print the
i-th walrus‘s displeasure: the number of other walruses that stand between him and the furthest from him younger walrus.

Sample Input

Input

6
10 8 5 3 50 45

Output

2 1 0 -1 0 -1 

Input

7
10 4 6 3 2 8 15

Output

4 2 1 0 -1 -1 -1 

Input

5
10 3 1 10 11

Output

1 0 -1 -1 -1 

线段树维护区间最小值， 循环一遍数组，每次找到比这个数小且在最右端的数，那么左边处理过的数可以更新成INF， 顺便更新最小值。那么查询之前， 先询问整个数组的最小是和当前值的比较，满足最小值小于当前值才有查询的必要，而且右边不成立就一定在左边。

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string>
#include <string.h>
#include <cmath>
#include <cctype>
#include <stack>
#include <queue>
#include <set>
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
using namespace std;
typedef __int64 LL;
const int maxn = 100005;
const int MAX = 0x3f3f3f3f;
int n, tg, b, k, a[maxn], mi[maxn << 2], ans[maxn];
void up(int o) {
    mi[o] = min(mi[o<<1], mi[o<<1|1]);
}
void build(int o, int l, int r) {
    if(l == r) {
        scanf("%d", &mi[o]);
        a[l] = mi[o];
        return;
    }
    int m = (l+r) >> 1;
    build(lson);
    build(rson);
    up(o);
}
void query(int o, int l, int r) {
    if(l == r) {
        ans[k++] = l - b - 1;
        return;
    }
    int m = (l+r) >> 1;
    if(mi[o<<1|1] < tg) query(rson);
    else query(lson);
}
void update(int o, int l, int r) {
    if(l == r) mi[o] = MAX;
    else {
        int m = (l+r) >> 1;
        if(b <= m) update(lson);
        else update(rson);
        up(o);
    }
}
int main()
{
    cin >> n;
    build(1, 1, n);
    for(int i = 1; i <= n; i++) {
        tg = a[i], b = i;
        update(1, 1, n);
        if(mi[1] >= tg) ans[k++] = -1;
        else query(1, 1, n);
    }
    for(int i = 0; i < k; i++) printf("%d ", ans[i]);
    return 0;
}