给定长度分别为 \(N\), \(M\) 的序列 \(A\), \(B\) .

序列 \(L\) 的构造方式如下：

-- lists are 1-indexed --

procedure generate_list(A, B, x):

    let n = length of A
    let m = length of B
    let L = an empty list

    for i from 1 to min(n, m - x), inclusive:
        for j from (i + x) to m, inclusive:
            Append (A[i]*B[j]) to the end of L

    return L

求出序列 \(L\) 中第 \(k\) 小的数字。

\(1\leq n,m\leq 2\times 10^5,1\leq x\leq m,1\leq |A_i|,|B_i|\leq 2\times 10^5,1\leq k\leq length(L)\)

关键是想到二分.

二分最后的答案 \(val\) .

从后往前枚举每个 \(a_i\) ，同时建立一个 bit ，维护当前 \(b\) 中数的范围 .

可以计算出有多少个数小于等于 \(val\) .

\(b\) 中每个数最多被插入依次，\(a\) 中每个数最多查询一次.

时间复杂度 : \(O(n\log m\log v)\)

空间复杂度 : \(O(n)\)

code

#include<bits/stdc++.h>
using namespace std;
const int add=2e5+10;
int n,m,X;
long long k;
int a[200010],b[200010];
int bit[400010];//从前往后
inline void upd(int i){
  while(i){
    bit[i]++;
    i-=i&-i;
  }
}
inline int qry(int i){
//  cout<<i<<endl;
  int res=0;
  while(i<=400010){
    res+=bit[i];
    i+=i&-i;
  }
  return res;
}
inline long long get1(long long x,long long y){
//  cout<<x<<","<<y<<endl;
  if(x>=0)return 1ll*x/y;
  else{
    if((-x)%y!=0)return -1ll*((-x)/y)-1;
    return -1ll*((-x)/y);
  }
}
inline long long get2(long long x,long long y){
  if(x>=0){
    if(x%abs(y)!=0)return -1ll*(x/(-y));
    return -1ll*(x/(-y));
  }
  else{
    if(x%abs(y)!=0)return 1ll*((-x)/(-y))+1;
    return 1ll*((-x)/(-y));
  }
}
bool check(long long tmp){
  memset(bit,0,sizeof(bit));
  long long sum=0;
  int cnt=0,lst=m;
  for(int i=n-1;i>=0;i--){
    for(int j=i+X;j<lst;j++){
      upd(b[j]+add);
      cnt++;
    }
    lst=min(lst,i+X);
  //  cout<<a[i]<<" "<<b[i+x]<<endl;
    if(a[i]>0){
      long long val=get1(tmp,a[i]);
    //  cout<<val<<endl;
      val=min(val,200008ll);
      val=max(val,-200008ll);
    //  cout<<val<<","<<cnt<<","<<qry(val+add+1)<<" ";
      sum+=cnt-qry(val+add+1);
    }
    else{
      long long val=get2(tmp,a[i]);
      val=min(val,200008ll);
      val=max(val,-200008ll);
    //  cout<<val<<endl;
    //  cout<<val<<","<<qry(val+add)<<" ";
      sum+=qry(val+add);
    }
  }
//  cout<<endl;
//  cout<<tmp<<" "<<sum<<endl;
  return sum>=k;
}
int main(){
  freopen("kth.in","r",stdin);
  freopen("kth.out","w",stdout);
  ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  cin>>n>>m>>X>>k;
  for(int i=0;i<n;i++)cin>>a[i];
  for(int i=0;i<m;i++)cin>>b[i];
  long long low=-40000000000,high=40000000000+1,ans=40000000000;
  while(low<high){
    long long mid=(low+high)>>1;
    if(check(mid)){
      ans=min(ans,mid);
      high=mid;
    }
    else{
      low=mid+1;
    }
  }
  cout<<ans<<endl;
  return 0;
}
/*inline? ll or int? size? min max?*/
/*
3 4 1 5
2 -3 1
3 1 -2 -1
*/