E n^2 可过...

简单优化一下即可...,幽默数据

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n;
ll a[214514];
ll s[214514],ss[214514];
ll cnt;
ll ans;

int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)
  {
  	cin>>a[i];
  	s[i]=s[i-1]+a[i];
  	ss[i]=ss[i-1]+(a[i]>0);
	}
	
	cnt=s[n]/3;
	if(cnt*3!=s[n])
	{
		cout<<0;
		return 0;
	}

	for(int i=1;i<=n;i++)
	{
		if(s[i]==cnt&&ss[i]>0)
		{
			for(int j=i+1;j<=n;j++)
			  if(ss[j]-ss[i]>0&&ss[n]-ss[j]>0)
			  	if(s[j]-s[i]==cnt)
			  		ans++;
		}
	}		
	
	cout<<ans;  		  
				
	return 0;
}