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：
1 #include<stdio.h>
2 #include<algorithm>
3 using namespace std;
4 const int MAXN = 5e5 + 10;
5
6 int n, m;
7 int a[MAXN], ANS[MAXN];
8
9 struct Tree
10 {
11 int val, l, r;
12 }tree[4 * MAXN];
13
14 void build(int k, int l, int r)
15 {
16 tree[k].l = l, tree[k].r = r;
17 if(l == r)
18 {
19 tree[k].val = a[l];
20 return ;
21 }
22 int mid = (l + r) / 2;
23 build(2 * k, l, mid);
24 build(2 * k + 1, mid + 1, r);
25 tree[k].val = max(tree[2 * k].val, tree[2 * k + 1].val);
26 }
27
28 int query(int k, int goal)
29 {
30 if(tree[k].l == tree[k].r)
31 return tree[k].l;
32 if(tree[2 * k + 1].val >= goal)
33 return query(2 * k + 1, goal);
34 else if(tree[2 * k].val >= goal)
35 return query(2 * k, goal);
36 else
37 return -1;
38 }
39
40 int main()
41 {
42 scanf("%d%d", &n, &m);
43 for(int i = 1; i <= n; i ++)
44 scanf("%d", &a[i]);
45 build(1, 1, n); //线段树建树
46 for(int i = 1; i <= n; i ++)
47 {
48 int flag = 0;
49 int ans = query(1, 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
50 if(ans == -1 || ans <= i)
51 ANS[i] = -1;
52 else if(ans > i)
53 ANS[i] = ans - i - 1;
54 }
55 printf("%d", ANS[1]);
56 for(int i = 2; i <= n; i ++)
57 printf(" %d", ANS[i]);
58 printf("\n");
59 return 0;
60 }
XKC's basketball team【线段树查询】
原文地址：https://www.cnblogs.com/yuanweidao/p/11493458.html