The Preliminary Contest for ICPC Asia Xuzhou 2019（补题）

E.XKC's basketball team

ps：

During the game misread the title, but have not found (a pain). That the number required to meet the requirements, there has been no idea, I did not expect is for the most long-distance, it is very easy.

solution：

Greedy from the back maintains a monotonically increasing sequence, because the role for i <j, a [i] <= a [j], then for i number before a [j] can completely replace a [i] is.

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=5e5+5;
 4 typedef long long ll;
 5 
 6 inline int read()
 7 {
 8     char ch=getchar();int x=0,f=0;
 9     while(ch<'0' || ch>'9') f|=ch=='-',ch=getchar();
10     while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
11     return f?-x:x;
12 }
13 
14 
15 int a[maxn];
16 int num[maxn];
17 vector <int > V;
18 int main(){
19     int n,m;n=read();m=read();
20     for(int i=1;i<=n;i++) a[i]=read();
21     V.push_back(n);num[n]=-1;
22     for(int i=n-1;i>=1;i--){
23         int tmp=a[i]+m;
24         int l=0;int r=V.size()-1;
25         int ans=-1;
26         while(l<=r){
27             int mid=(l+r)/2;
28             if(a[V[mid]]>=tmp) 
29             {
30                 ans=max(ans,V[mid]); 
31                 r=mid-1;
32             }
33             else{
34                 l=mid+1;
35             } 
36         }
37         if(ans==-1) num[i]=ans;
38             else num[i]=ans-i-1;
39         if(a[i]>a[V[V.size()-1]]) V.push_back(i);
40         
41     }
42     for(int i=1;i<=n;i++){
43         if(i==1) cout <<num[i];
44         else cout <<" "<<num[i];
45     }    
46     return 0;
47 }