Gym

题意：good element的定义是a[i]在1~i-1中任取三个数（可以重复）的和能等于a[i]

思路：vis[x]标记一下任两个数的和，处理a[i]时枚举1~i-1判断vis[a[i] - a[j]]是否被标记

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#define LL long long
#define eps 1e-8
#define INF 0x3f3f3f3f
#define MAXN 200005
using namespace std;
int a[5005];
bool vis[400005];
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // OPEN_FILE
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++){
        scanf("%d", &a[i]);
    }
    int ans = 0;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= i - 1; j++){
            if(!vis[a[i] - a[j] + MAXN]) continue;
            ans++;
            break;
        }
        for(int j = 1; j <= i; j++){
            vis[a[i] + a[j] + MAXN] = true;
        }
    }
    printf("%d
", ans);
}