题目传送门：https://codeforces.com/problemset/problem/91/B
题目大意：
给一个长度为(n)的序列(A)，记 (B_i=j-i,j=max{j|i<jand A_j<A_i})，若不存在这样的 (j)，则记 (B_i=-1)，求(B)的值
考虑用树状数组维护，倒序查询可以保证 (i<j) 这一限制条件，以权值作为下标，位置作为权值，便可以求出 (A_j<A_i) 中最大的 (j)
/*program from Wolfycz*/
#include<map>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Fi first
#define Se second
#define ll_inf 1e18
#define MK make_pair
#define sqr(x) ((x)*(x))
#define pii pair<int,int>
#define int_inf 0x7f7f7f7f
#define lowbit(x) ((x)&(-x))
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
static char buf[1000000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template<typename T>inline T frd(T x){
int f=1; char ch=gc();
for (;ch<'0'||ch>'9';ch=gc()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
template<typename T>inline T read(T x){
int f=1; char ch=getchar();
for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
inline void print(int x){
if (x<0) putchar('-'),x=-x;
if (x>9) print(x/10);
putchar(x%10+'0');
}
const int N=1e5;
int A[N+10],Ans[N+10],list[N+10],Tree[N+10],n;
void Add(int x,int v){for (;x<=n;x+=lowbit(x)) Tree[x]=max(Tree[x],v);}
int Query(int x){
int res=0;
for (;x;x-=lowbit(x)) res=max(res,Tree[x]);
return res;
}
int main(){
n=read(0);
for (int i=1;i<=n;i++) A[i]=list[i]=read(0);
sort(list+1,list+1+n);
int T=unique(list+1,list+1+n)-list-1;
for (int i=1;i<=n;i++) A[i]=lower_bound(list+1,list+1+T,A[i])-list;
for (int i=n;i>=1;i--){
Ans[i]=Query(A[i]-1);
Ans[i]=!Ans[i]?-1:Ans[i]-i-1;
Add(A[i],i);
}
for (int i=1;i<=n;i++) printf("%d%c",Ans[i],i==n?'
':' ');
return 0;
}