The Preliminary Contest for ICPC Asia Xuzhou 2019 E. XKC's basketball team

Topic links: https://nanti.jisuanke.com/t/41387

Thinking: We need to move forward to maintain an ascending sequence from the rear.

Because: We want wi + m <= wj, j to take the maximum, i.e. the farthest from the j i, wi each index requires a determination is not greater than w (i + 1), to complete the sequence of increments maintain.

Why can better understand the code inside to maintain an ascending sequence

---

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <string>
#include <map>
#include <cmath>
using namespace std;
 
typedef long long LL;
#define inf 1e9
#define rep(i,j,k) for(int i = (j); i <= (k); ++i)
#define rep__(i,j,k) for(int i = (j); i < (k); ++i)
#define per(i,j,k) for(int i = (j); i >= (k); --i)
#define per__(i,j,k) for(int i = (j); i > (k); --i) 

const int N = (int)5e5 + 10;
int arr[N];
int ans[N];

struct node{
    int index;
    int w;
    
    void set(int x,int y){
        index = x;
        w = y;
    }
}que[N];

int main(){

    int n,m;
    scanf("%d%d",&n,&m);
    rep(i,1, n-) Scanf ( " % D " , ARR + I); 

    int L, R & lt, MID, A;
     int len = 0 ;
     // last process to 
    que [++ len]. SET (n-, ARR [n-]) ; 
    ANS [n-] = - . 1 ;
     //
     per (I, N- . 1 , . 1 ) { 

        // than before are large 
        IF (ARR [I] + m> que [len] .W) { 
            ANS [I] = - . 1 ;
             IF (ARR [I]> que [len] .W) 
                que [ ++ len]. SET (I, ARR [I]); 
        } 
        //Smaller than the first one 
        the else  IF (ARR [I] + m <= que [ . 1 ] .W) { 
            ANS [I] = que [ . 1 ] .index - I - . 1 ;
             // the printf ( "% D IS the this ANS IS% D Test \ n-", I, ANS [I]); 
        }
         // in the middle answer 
        the else { 
            L = . 1 ; 
            R & lt = len;
             the while (L <= R & lt) { 
                mID = (L + R & lt)> > . 1 ;
                 IF (ARR [I] + m <=  que [MID] .W) {
                    A = que [MID] .index;
                    r = mid - 1;
                }
                else l = mid + 1;
            }
  //          printf("arr[%d] = index %d - index %d - 1\n",i,A,i);
            ans[i] = A - i -1;
        }        
    }

    // rep(i,1,ll) printf("%d ",que[i].index);
    // printf("\n");

    printf("%d",ans[1]);
    rep(i,2,n) printf(" %d",ans[i]);
    printf("\n");

    getchar(); getchar();
    return 0;
}