#include <bits/stdc++.h>
#define endl '\n'
using namespace std;
typedef pair<int, int> PII;
using ll = long long;
using ULL = unsigned long long;
const int N = 1e7 + 5;

int n, a[N];
inline void solve() {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    int m; cin >> m;
    for (; m-- ;) {
        int l, r, k;
        cin >> l >> r >> k;
         vector<int> b;  
        b.reserve(r - l + 1);  
        for (int i = l; i <= r; i++) {
            b.push_back(a[i]);
        }
        nth_element(b.begin(), b.begin() + (k-1), b.end(), greater<int>());
        cout << b[k-1] << endl;
    }
}

int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int _ = 1;
    //int _; cin >> _;
    while (_--) solve();
    return 0;
}

一开始并非从大到小 需要先选择l和r,再进行sort排序

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
struct number{
	int l;
	int r;
	int K;
	int b[10005];
	int ret;
}pp[10005];
int main()
{
	int n;
	int a[1005];
	cin >> n;
	for(int i = 1 ;i <= n ;i++){
		cin >> a[i];
	}
    int m;
	cin >> m;
	for(int i = 1 ;i <= m ;i++){
		cin >> pp[i].l >>pp[i].r >> pp[i].K;
		int k = 0;
		for(int j = pp[i].l ;j <= pp[i].r ;j++){
			k++;
			pp[i].b[k] = a[j]; 
		}
		sort(pp[i].b + 1 ,pp[i].b + 1 + k);
		pp[i].ret = pp[i].b[k+1-pp[i].K];
	}
	for(int i = 1 ;i <= m ;i++){
		cout << pp[i].ret << "\n";
	}
	return 0;
}