#include <bits/stdc++.h>
using namespace std;
struct F
{
    int id,score;
}a[5005];
bool cmp(F x,F y)//简单排序
{   
    if(x.score != y.score)
    {
        return x.score > y.score;
    }
    else
    {
        return x.id < y.id;
    }
}
int main()
{
    int n,m;
    cin >> n >> m;
    int q = floor(m * 1.5);
    for(int i = 1; i <= n; i++)
    {
        cin >> a[i].id >> a[i].score;
    }
    sort(a + 1,a + n + 1,cmp);
    int op = a[q].score,num = 0;
    for(int i = 1; i <= n; i++)
    {
        if(a[i].score >= op)//统计人数
        {
            num++;
        }
    }
    cout << op << " " << num << endl;
    for(int i = 1; i <= num; i++)
    {
        cout << a[i].id << " " << a[i].score << endl;
    }
    return 0;
}

唯一需要注意的就是成绩相同并不是并列名次，只是和面试成绩线相同的时候才算并列哈。

上代码！

#include <bits/stdc++.h>
using namespace std;
int n, m, t, k[5005], s[5005], r[5005];
bool cmp(int a, int b)
{
    if (s[a] != s[b])
        return s[a] > s[b];
    return k[a] < k[b];
}
int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; ++i)
    {
        cin >> k[i] >> s[i];
        r[i] = i;
    }
    t = m = m * 1.5;
    sort(r + 1, r + n + 1, cmp);
    while (s[r[t]] >= s[r[m]])
        t++;
    cout << s[r[m]] << " " << t - 1 << endl;
    for (int i = 1; i < t; ++i)
        cout << k[r[i]] << " " << s[r[i]] << endl;
    return 0;
} //代码已AC

题解

学过我的题解的都知道，这是一个结构体排序，然后计算一下分数线，统计一下人数，输出，完成。

#include <bits/stdc++.h>
using namespace std;
int n, m, pass, s = 0;
struct mark
{
    int a, h;
};
mark mn[5000];
bool cmp(mark x, mark y)
{
    if (x.a > y.a) return true;
	if (x.a == y.a && x.h < y.h) return 1;
    return 0;  
}
int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
        scanf("%d%d", &mn[i].h, &mn[i].a);
    sort(mn + 1, mn + n + 1, cmp);
	pass = mn[m * 3 / 2].a;
    for (int i = 1; i <= n; i++)
    	if (mn[i].a >= pass) s++;
	printf("%d %d\n", pass, s);
	for (int i = 1; i <= s; i++)
        printf("%d %d\n", mn[i].h, mn[i].a);
    return 0;
}

简单的没话说。

$AC$ $Code$

#include <bits/stdc++.h>
using namespace std;
int n, m, pass, s = 0;
struct mark
{
    int a, h;
};
mark mn[5000];
bool cmp(mark x, mark y)
{
    if (x.a > y.a) return true;
	if (x.a == y.a && x.h < y.h) return 1;
    return 0;  
}
int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
        scanf("%d%d", &mn[i].h, &mn[i].a);
    sort(mn + 1, mn + n + 1, cmp);
	pass = mn[m * 3 / 2].a;
    for (int i = 1; i <= n; i++)
    	if (mn[i].a >= pass) s++;
	printf("%d %d\n", pass, s);
	for (int i = 1; i <= s; i++)
        printf("%d %d\n", mn[i].h, mn[i].a);
    return 0;
}

#include<iostream>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
struct node{//定义结构体 
	int num;
	int grade;
};
bool cmp(node a,node b){//编写cmp函数，实现成绩从大到小排序，成绩相同，编号从小到大 
	if(a.grade!=b.grade)
	return a.grade>b.grade;
	else
	return a.num<b.num;
}
int n,m;
node a[5000]; 
int mian(){
    cin>>n>>m;
    int i,temp,j;
    for(i=0;i<n;i++){
    	cin>>a[i].num>>a[i].grade;
	}
	sort(a,a+n,cmp);//排序 
	temp=m*1.5;
	cout<<a[temp-1].grade<<" ";
	for(i=temp;a[i].grade==a[temp-1].grade;i++);//精髓for循环 
	
		cout<<i<<endl;
		for(j=0;j<i;j++){
			cout<<a[j].num<<" "<<a[j].grade<<endl;
		}
	
    
    return -1;
}

PS:请勿复制此题解，其中有小错误，建议自己理解