Gym - 100203G Good elements 水+模拟

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[];
 bool vis[];
 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 = ; i <= n; i++){
         scanf("%d", &a[i]);
     }
     int ans = ;
     for(int i = ; i <= n; i++){
         for(int j = ; j <= i - ; j++){
             if(!vis[a[i] - a[j] + MAXN]) continue;
             ans++;
             break;
         }
         for(int j = ; j <= i; j++){
             vis[a[i] + a[j] + MAXN] = true;
         }
     }
     printf("%d\n", ans);
 }