描述

Given a sequence including N integers, we define the ith element "good" if it equals the sum of three elements strictly before the position i (an element can be used more than once).
Please tell me the number of good elements int this sequence?

輸入

The first line is an integer N (1 <= N <= 5000), and the next line has N integers indicating the elements of the sequence, each element is between -100000 and 100000.

輸出

Output an integer indicating the number of good elements in the sequence.

樣例輸入

2
1 3

5

1 2 3 4 5

樣例輸出

1

3

計算出前面每兩項之和， 用要判斷的那一個數 減去前面的， 如果它們的結果存在的話 （這裏用1標記）， 就說明是好數。

#include <iostream>
using namespace std;
int num[400005]={0};             //存兩數之和是否存在，有負數 ，所以要開大點
int a[5005];
int main()
{
    
 
int n,m,i,j,flag,ans=0;
    cin>>n;
    for(i=1;i<=n;i++)
    scanf("%d",&a[i]);
    for(i=1;i<=n;i++){
        for(j=1;j<i;j++){
            num[a[i-1]+a[j]+200000]=1;  
        }
        for(j=1;j<i;j++){
            if(num[a[i]-a[j]+200000]==1){     
                   ans++;
                
 
break;
            }
        }
    }
    cout<<ans<<endl;
}

Good elements