Codeforces Gym 100203G G

G - Good elements
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87954#problem/G

Description

You are given a sequence A consisting of N integers. We will call the i-th element good if it equals the sum of some three elements in positions strictly smaller than i (an element can be used more than once in the sum).

How many good elements does the sequence contain?

Input

The first line of input contains the positive integer N (1 ≤ N ≤ 5000), the length of the sequence A.

The second line of input contains N space-separated integers representing the sequence A ( - 10⁵ ≤ A_i ≤ 10⁵).

Output

The first and only line of output must contain the number of good elements in the sequence.

Sample Input

2
1 3

Sample Output

1

HINT

题意

      给你一个序列，任意一个数只要等于在它前面的任意三个数之和（可以相同）则称为good数

题解：

      有时候能用数组尽量用数组，set还是有祸害的

代码：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <queue>
#include <typeinfo>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
inline ll read()
{
    ll x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
//***************************
int s[5000000];
 int main()
 {
     int ans=0;
     int a[50004];
     int n=read();
     for(int i=1;i<=n;i++)
     {
         scanf("%d",&a[i]);
         for(int j=1;j<i;j++)
         {
             if(s[a[i]-a[j]+500000])
             {
                 ans++;
                 break;
             }
         }
         for(int j=1;j<i;j++)
         {
             s[a[i]+a[j]+500000]=1;
         }
         s[a[i]+a[i]+500000]=1;
     }
     cout<<ans<<endl;
     return 0;
 }