1. We can first consider the violent approach, first break the ring into a column, and then judge one by one to find the prefix sum, and then judge one by one whether all are >=0, and the complexity is O(n^2)
2. The monotonic queue maintains the least The prefix sum, for each k+n-1, we use a monotonic queue to maintain the minimum value of k–k+n-1, and subtract s[k-1] to determine whether it is legal. If the smallest is already legal, then it must be legal
#include<bits/stdc++.h>
#define maxn 2000010
using namespace std;
int n,m;
long long a[maxn],q[maxn],sum[maxn];
int head=1,tail;
inline long long read()
{
long long res=0,f=1;char ch=getchar();
while(!isdigit(ch)){
if(ch=='-')f=-f;ch=getchar();}
while(isdigit(ch)){
res=(res<<1)+(res<<3)+(ch&15);ch=getchar();}
return res*f;
}
int ans;
int main()
{
cin>>n;
for(int i=1;i<=n;i++)a[i]=read(),a[i+n]=a[i];
for(int i=1;i<n*2;i++)sum[i]+=sum[i-1]+a[i];
for(int i=1;i<n*2;i++)
{
if(q[head]<=i-n)head++;
// cout<<q[head]<<' ';
while(head<=tail&&(sum[i]<=sum[q[tail]]))tail--;
q[++tail]=i;
if(i>=n&&sum[q[head]]-sum[i-n]>=0)ans++;
}
cout<<ans;
return 0;
}