XKC , the captain of the basketball team , is directing a train of nnn team members. He makes all members stand in a row , and numbers them 1⋯n1 cdots n1⋯n from left to right.
The ability of the iii-th person is wiw_iwi , and if there is a guy whose ability is not less than wi+mw_i+mwi+m stands on his right , he will become angry. It means that the jjj-th person will make the iii-th person angry if j>ij>ij>i and wj≥wi+mw_j ge w_i+mwj≥wi+m.
We define the anger of the iii-th person as the number of people between him and the person , who makes him angry and the distance from him is the longest in those people. If there is no one who makes him angry , his anger is −1-1−1 .
Please calculate the anger of every team member .
Input
The first line contains two integers nnn and m(2≤n≤5∗105,0≤m≤109)m(2leq nleq 5*10^5, 0leq m leq 10^9)m(2≤n≤5∗105,0≤m≤109) .
The following line contain nnn integers w1..wn(0≤wi≤109)w_1..w_n(0leq w_i leq 10^9)w1..wn(0≤wi≤109) .
Output
A row of nnn integers separated by spaces , representing the anger of every member .
样例输入
6 1 3 4 5 6 2 10
样例输出
4 3 2 1 0 -1
解题思路：单调栈+二分 从后往前维护一个单调递增的栈 然后二分查找即可！
1 #include <bits/stdc++.h> 2 using namespace std; 3 int n,m; 4 const int maxn=5e5+5; 5 int arr[maxn]; 6 int sta[maxn]; 7 int top=0; 8 map<int,int> ma; 9 10 int Binary(int left,int right,int value){ 11 int ans=-1; 12 while(left<=right){ 13 int mid=left+right>>1; 14 if(arr[sta[mid]]>=value) ans=mid,right=mid-1; 15 else left=mid+1; 16 } 17 return ans; 18 } 19 20 int main(){ 21 ios::sync_with_stdio(false); 22 cin>>n>>m; 23 for(int i=1;i<=n;i++) cin>>arr[i]; 24 int top=1; 25 sta[top]=n;ma[n]=-1; 26 for(int i=n-1;i>=1;i--){ 27 if(arr[i]>arr[sta[top]]-m) ma[i]=-1; 28 else{ 29 int flag=Binary(1,top,arr[i]+m); //找大于等于这个数的最左边的位置 30 // cout << arr[sta[flag]] << endl; 31 ma[i]=sta[flag]-i-1; 32 } 33 if(arr[i]>arr[sta[top]]) sta[++top]=i; 34 } 35 for(int i=1;i<=n;i++){ 36 printf("%d%c",ma[i],i==n?' ':' '); 37 } 38 return 0; 39 } 40 41 //10 1 42 //3 4 5 6 2 20 10 30 40 50