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
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;
}