D. Non-zero Segments

D. Non-zero Segments

过程：手玩样例发现，前缀和中出现相同值即为区间和为0

由前缀和公式\(S_i = S_{i - 1} + A_i\)可以知道，区间\([l,r]\)满足题意即为\(S_r = S_{l-1}\)，对于每个这样的区间插入个巨大的数，让区间\([1, r-1]\)的前缀和都作废，注意对0处理

even such a huge that they can't be represented in most standard programming languages

维护是否有相同的前缀和可以用\(\text{set}\)

更多细节见代码

ll n, s[N], ans; 
set<ll> S;
void solve()
{   
	cin>>n;
	for(int i = 1; i <= n; i++)
		cin>>s[i];
	S.insert(0);
	for(int i = 1; i <= n; i++)
	{
		s[i] += s[i - 1];
		if(S.count(s[i]))
		{	
			ans++;
			S.clear();
			S.insert(0);
			s[i] -= s[i - 1];
		}
		S.insert(s[i]);
	}
	cout<<ans<<endl;

    return;
}