CF91B
题意是十分的简单
自己因为Min函数敲错 还有build 时候忘记up 心疼自己的时间
/*
91B
if you can't see the repay
Why not just work step by step
rubbish is relaxed
to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define lc (rt<<1)
#define rc (rt<<11)
#define mid ((l+r)>>1)
typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod = (int)1e9+7;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
const int MAX_N = 100025;
int minn[MAX_N<<2],ans[MAX_N],arr[MAX_N];
int Min(int a,int b) {if(a<b) return a; return b;}
void up(int rt)
{
minn[rt] = Min(minn[rt<<1],minn[rt<<1|1]);
}
void build(int rt,int l,int r)
{
if(l==r)
{
minn[rt] = arr[l];
return;
}
build(rt<<1,l,mid);
build(rt<<1|1,mid+1,r);
up(rt);
}
int query(int rt,int l,int r,int x,int y,int v)
{
if(minn[rt]>v) return -1;
if(l==r)
{
return l;
}
int ans =-1;
if(minn[rt<<1|1]<v&&y>mid) ans = query(rt<<1|1,mid+1,r,x,y,v);
if(ans==-1&&x<=mid) ans = query(rt<<1,l,mid,x,y,v);
return ans;
}
int main()
{
//ios::sync_with_stdio(false);
//freopen("a.txt","r",stdin);
//freopen("b.txt","w",stdout);
int n;
scanf("%d",&n);
for(int i = 1;i<=n;++i) scanf("%d",&arr[i]);
build(1,1,n);
for(int i = 1;i<n;++i)
{
int tmp = query(1,1,n,i+1,n,arr[i]);
ans[i] = (tmp==-1?-1:tmp-i-1);
}
ans[n] = -1;
for(int i = 1;i<=n;++i)
i==n?printf("%d\n",ans[i]):printf("%d ",ans[i]);
//fclose(stdin);
//fclose(stdout);
//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
return 0;
}