The Preliminary Contest for ICPC Asia Xuzhou 2019 徐州网络赛 XKC's basketball team

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(2leq nleq 5*10^5, 0leq m leq 10^9)m(2≤n≤5∗105,0≤m≤109) .

The following  line contain nn integers w_1..w_n(0leq 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

队友做的，没大有思路。

#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
#define LL long long
#define MAXN 500100
struct node
{
    LL x;
    LL y;
} a[MAXN*2];
LL b[MAXN],c[MAXN];
bool cmp(node u,node v)
{

    return u.x<v.x;
}
int main()
{
    LL n,m,i,j;
    scanf("%lld%lld",&n,&m);
    for(i=1; i<=n; i++)
    {
        scanf("%lld",&a[i].x);
        a[i].y=i;
        b[i]=a[i].x;
    }
    j=n;
    sort(a+1,a+n+1,cmp);
    bool flag=false;
    for(i=1; i<=n; i++)
    {
        while(b[j]<a[i].x+m)
        {
            if(j==0)
            {
                flag=true;
                break;
            }
            j--;
        }
        if(flag)
            c[a[i].y]=-1;
        else{
            if(j<=a[i].y) c[a[i].y]=-1;
        else
            c[a[i].y]=j-a[i].y-1;
            }
    }
    for(i=1; i<n; i++)
        printf("%lld ",c[i]);
    printf("%lld",c[n]);
    return 0;
}