题目解析
给定n个已知关注的
贼给定m个 我给点赞的信息
假如我点赞的次数大于平均数 则为悄悄关注的
输出悄悄关注的人时，升序排序
解题思路
利用set的性质去判断是否有该元素
然后再去循环判断是否大于平均值
然后 将找出来的名字排序输出即可
注: java最后一个样例我过不去
代码
import java.io.*;
import java.math.*;
import java.util.*;
public class Main
{
public static void main(String[] args) throws IOException
{
int n = sc.nextInt();
TreeSet<String> tr = new TreeSet<String>();
while (n-- > 0)
tr.add(sc.next());
int m = sc.nextInt();
String s[] = new String[m + 10];
int shu[] = new int[m + 10];
double ans = 0;
for (int i = 1; i <= m; i++)
{
s[i] = sc.next();
shu[i] = sc.nextInt();
ans += shu[i];
}
ans /= m;
ArrayList<String> ar = new ArrayList<String>();
for (int i = 1; i <= m; i++)
{
if (shu[i] > ans && !tr.contains(s[i]))
ar.add(s[i]);
}
if (ar.size() == 0)
out.println("Bing Mei You");
else
{
Collections.sort(ar);
for (String i : ar)
out.println(i);
}
out.flush();
out.close();
}
static Scanner sc = new Scanner(System.in);
static PrintWriter out = new PrintWriter(System.out);
}
c++
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
using namespace std;
set<string> se;
vector<string> ve;
const int N= 1e4;
string s[N + 10];
int shu[N + 10];
int main()
{
int n; cin >> n;
while(n -- > 0)
{
string res; cin >> res;
se.insert(res);
}
int m; cin >> m;
double ans = 0;
for(int i = 1; i <= m; i ++)
{
cin >> s[i] >> shu[i];
ans += shu[i];
}
ans /= m;
for(int i = 1; i <= m; i ++)
{
if(shu[i] > ans && !se.count(s[i]))
ve.push_back(s[i]);
}
if(ve.size() == 0)
cout << "Bing Mei You";
else
{
sort(ve.begin(), ve.end());
for(string i : ve)
cout << i << endl;
}
return 0;
}