算法

(冒泡排序) $O(n^2)$

开两个数组，一个数组存编号，另一个数组存成绩，然后用冒泡排序模拟一下就行了

C++ 代码

#include <iostream>

using namespace std;

const int N  = 5010;

#define rep(i, n) for (i = 1; i <= n; i++)

int a[N], b[N];
int n, m;


int main() {
    cin >> n >> m;
    int i;
    rep(i, n) cin >> a[i] >> b[i];

    int j;
    rep(i, n - 1) {
        rep(j, n - i) {
            if (b[j] < b[j + 1] || b[j] == b[j + 1] && a[j] > a[j + 1]) {
                swap(a[j], a[j + 1]);
                swap(b[j], b[j + 1]);
            }
        }
    } 

    m *= 1.5;
    cout << b[m];

    int cnt = 0;
    rep(i, n) {
        if (b[i] >= b[m]) cnt++;
    }
    cout << " " << cnt << endl;

    rep(i, cnt) cout << a[i] << " " << b[i] << endl;

    return 0;
}