XKC's basketball team【线段树查询】

XKC , the captain of the basketball team , is directing a train of nn team members. He makes all members stand in a row , and numbers them 1 \cdots n1⋯n from left to right.

The ability of the ii-th person is w_iwi​ , and if there is a guy whose ability is not less than w_i+mwi​+m stands on his right , he will become angry. It means that the jj-th person will make the ii-th person angry if j>ij>i and w_j \ge w_i+mwj​≥wi​+m.

We define the anger of the ii-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 .

Please calculate the anger of every team member .

Input

The first line contains two integers nn and m(2\leq n\leq 5*10^5, 0\leq m \leq 10^9)m(2≤n≤5∗105,0≤m≤109) .

The following  line contain nn integers w_1..w_n(0\leq w_i \leq 10^9)w1​..wn​(0≤wi​≤109) .

Output

A row of nn integers separated by spaces , representing the anger of every member .

样例输入复制

6 1
3 4 5 6 2 10

样例输出复制

4 3 2 1 0 -1

题目大意：
1.给出一个长度为n的数列，给出m的值。问在数列中从右到左第一个比 arr[i] + m 大的值所在位置与 arr[i] 的所在位置之间夹着多少个数。
解题思路：
1.用线段树来记录区间最大值，优先从右子树开始查询是否存在大于 arr[i] + m的值，返回下标即该值在原数列中的位置，即可求得答案。
2.重要的是查询的值必须是在 i 的右边，否则输出 -1
代码如下：

 #include<stdio.h>
 #include<algorithm>
 using namespace std;
 const int MAXN = 5e5 + ;

 int n, m;
 int a[MAXN], ANS[MAXN];

 struct Tree
 {
     int val, l, r;
 }tree[ * MAXN];

 void build(int k, int l, int r)
 {
     tree[k].l = l, tree[k].r = r;
     if(l == r)
     {
         tree[k].val = a[l];
         return ;
     }
     int mid = (l + r) / ;
     build( * k, l, mid);
     build( * k + , mid + , r);
     tree[k].val = max(tree[ * k].val, tree[ * k + ].val);
 }

 int query(int k, int goal)
 {
     if(tree[k].l == tree[k].r)
         return tree[k].l;
     if(tree[ * k + ].val >= goal)
         return query( * k + , goal);
     else if(tree[ * k].val >= goal)
         return query( * k, goal);
     else
         return -;
 }

 int main()
 {
     scanf("%d%d", &n, &m);
     for(int i = ; i <= n; i ++)
         scanf("%d", &a[i]);
     build(, , n); //线段树建树
     for(int i = ; i <= n; i ++)
     {
         int flag = ;
         int ans = query(, a[i] + m); //由右边开始查询，返回右边第一个大于 a[i] + m值的位置
         if(ans == - || ans <= i)
             ANS[i] = -;
         else if(ans > i)
             ANS[i] = ans - i - ;
     }
     printf("%d", ANS[]);
     for(int i = ; i <= n; i ++)
         printf(" %d", ANS[i]);
     printf("\n");
     return ;
 }