Codeforces Beta Round #75 (Div. 1 Only) B. Queue 二分

B. Queue

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

codeforces.com/problemset/problem/91/B

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 ai.

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 ai > aj. 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 ai (1 ≤ ai ≤ 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

6
10 8 5 3 50 45

Sample Output

2 1 0 -1 0 -1

HINT

 

题意

给你一个数列，让你找到最右边比这个数小的数的位置，如果没有就输出-1

题解：

对与某一位i,假设最右边比他大的数为j，那么a[i]>a[j]，然后将min{a[j+1],a[j+2],...,a[n]}记作min[j+1], 则a[i]<=min[j+1],

不难想到不等式：　min[j]<=a[j]<a[i]<=min[j+1]

这就说明min数组单增，然后在单增序列里面找一个值。

这不就是二分吗？然后就没了。

 

代码

 

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 100050
int n,a[N],mi[N];
template<typename T>void read(T&x)
{
  ll k=0; char c=getchar();
  x=0;
  while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
  if (c==EOF)exit(0);
  while(isdigit(c))x=x*10+c-'0',c=getchar();
  x=k?-x:x;
}
void read_char(char &c)
{while(!isalpha(c=getchar())&&c!=EOF);}
int ef(int x,int l,int r)
{
  if (l==r)return l;
  int mid=(l+r+1)>>1;
  if (x>mi[mid])
    return ef(x,mid,r);
  else return ef(x,l,mid-1);
}
int main()
{
#ifndef ONLINE_JUDGE
  freopen("aa.in","r",stdin);
#endif
  read(n);
  for(int i=1;i<=n;i++)read(a[i]);
  mi[n]=a[n];
  for(int i=n-1;i>=1;i--)mi[i]=min(mi[i+1],a[i]);
  for(int i=1;i<=n;i++)
    {
      int ans=ef(a[i],i,n)-i-1;
      printf("%d ",ans);
    }
}