CodeForces 91B Queue【线段树】

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本文介绍了一个利用线段树数据结构解决特定队列问题的方法。该问题涉及计算队列中每只海象因前方年轻海象而产生的不满程度。通过维护区间最小值，算法能高效地处理这一问题。

Queue

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 91B

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. Thedispleasure 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，每次找到比这个数小且在最右端的数，将左边处理过的数可以更新成INF并更新最小值。那么查询之前， 先询问整个数组的最小是和当前值的比较，当前值大于最新最小值才有查询的必要，而且右边不成立就一定在左边。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<climits>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<algorithm>
using namespace std;
#define rep(i,j,k)for(i=j;i<k;i++)
#define per(i,j,k)for(i=j;i>k;i--)
#define MS(x,y)memset(x,y,sizeof(x))
#define max(a,b) a>b?a:b
#define min(a,b) a<b?a:b
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long LL;
const int INF=0x7ffffff;
#define lson rt<<1, l, m
#define rson rt<<1|1, m+1, r
const int maxn = 100005;
const int MAX = 0x3f3f3f3f;
int n,i,j,k,b;
int cnt;
int a[maxn],mi[maxn<<2],ans[maxn];

void up(int rt)
{
    mi[rt]=min(mi[rt<<1],mi[rt<<1|1]);
}

void build(int rt ,int l,int r)
{
    if(l==r){
        scanf("%d",&mi[rt]);
        a[l]=mi[rt];
        return;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    up(rt);
}

void query(int rt,int l,int r)
{
    if(l==r){
        ans[k++]=l-b-1;
        return;
    }
    int m=(l+r)>>1;
    if(mi[rt<<1|1]<cnt)query(rson);
    else query(lson);
}

void update(int rt,int l,int r)
{
    if(l==r)mi[rt]=MAX;
    else {
        int m=(l+r)>>1;
        if(b<=m)update(lson);
        else update(rson);
        up(rt);
    }
}

int main()
{
    scanf("%d",&n);
    build(1,1,n);
    for(int i=1;i<=n;i++){
        cnt=a[i],b=i;
        update(1,1,n);
        if(mi[1]>=cnt)ans[k++]=-1;
        else query(1,1,n);
    }
    for(i=0;i<k;i++)printf("%d ",ans[i]);
    return 0;
}