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

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <fstream>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include <deque>
 7 #include <vector>
 8 #include <queue>
 9 #include <string>
10 #include <cstring>
11 #include <map>
12 #include <stack>
13 #include <set>
14 #define LL long long
15 #define eps 1e-8
16 #define INF 0x3f3f3f3f
17 #define MAXN 200005
18 using namespace std;
19 int a[5005];
20 bool vis[400005];
21 int main()
22 {
23 #ifndef ONLINE_JUDGE
24     freopen("in.txt", "r", stdin);
25     //freopen("out.txt", "w", stdout);
26 #endif // OPEN_FILE
27     int n;
28     scanf("%d", &n);
29     for(int i = 1; i <= n; i++){
30         scanf("%d", &a[i]);
31     }
32     int ans = 0;
33     for(int i = 1; i <= n; i++){
34         for(int j = 1; j <= i - 1; j++){
35             if(!vis[a[i] - a[j] + MAXN]) continue;
36             ans++;
37             break;
38         }
39         for(int j = 1; j <= i; j++){
40             vis[a[i] + a[j] + MAXN] = true;
41         }
42     }
43     printf("%d\n", ans);
44 }

 

posted on 2015-08-16 00:53  张济  阅读(184)  评论(0)  编辑  收藏  举报

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