K-th decimal quick sorting variant

Topic link

Handwritten quick sorting, each time sorting judges the position of the benchmark m,, if m> k, the kth decimal is at
[l,m-1], if m <k, the kth decimal is at [m + 1,r], Otherwise, m == k, and num[m] is the kth decimal.

If the amount of data is too large, you need to write a quick read by yourself, otherwise the reading process may exceed the limit

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll num[6000000];
ll T,n,k;

void read(ll &x){
    
    
    char c;
    ll flag = 1;
    x = 0;
    while((c = getchar()) && (c < '0' || c > '9'))if(c =='-')flag = -1;
    x = c - '0';
    while((c = getchar()) && (c >= '0' && c <= '9')) x = x*10 + c -'0';
    x*=flag;
}

ll qsort(ll l,ll r){
    
    
    
    if(l == r)   return num[l];
    ll m = l,head = l,rear = r;
    num[0] = num[m];
    
    while(head < rear){
    
    
        while(num[0] <= num[rear] && rear > m)    rear--;
             num[m] = num[rear];// '>='两个数一样不用交换 rear > m 防止越界
             m = rear;
        
        while(num[0] >= num[head] && head < m)    head++;
            num[m] = num[head];
            m = head;
    }
    
    num[m] = num[0];
   
    if(m < k)
        return qsort(m + 1,r);
    
    else if(m > k)
        return qsort(l,m-1);
    
    return num[k];
    
}

int main(){
    
    
   
    read(T);
    for(int t = 0;t < T;t++){
    
    
        read(n); read(k);
        for(int i = 1;i <= n;i++)
            read(num[i]);
        cout << qsort(1,n) << endl;
    }
    return 0;
}