Funny Positive Sequence （思维+前缀）

There are n integers a ₁,a ₂,…,a _n-1,a _n in the sequence A, the sum of these n integers is larger than zero. There are n integers b ₁,b ₂,…,b _n-1,b _n in the sequence B, B is the generating sequence of A and bi = a ₁+a ₂,+…+a _i (1≤i≤n). If the elements of B are all positive, A is called as a positive sequence.

We left shift the sequence A 0,1,2,…,n-1 times, and get n sequences, that is showed as follows:

A(0): a₁,a₂,…,a_n-1,a_n

A(1): a₂,a₃,…,a_n,a₁

…

A(n-2): a_n-1,a_n,…,a_n-3,a_n-2

A(n-1): a_n,a₁,…,a_n-2,a_n-1

Your task is to find out the number of positive sequences in the set { A(0), A(1), …, A(n-2), A(n-1) }.

Input

The first line of the input contains an integer T (T <= 20), indicating the number of cases. Each case begins with a line containing one integer n (1 <= n <= 500,000), the number of elements in the sequence. The next line contains n integers ai(-2,000,000,000≤ai≤2,000,000,000,1≤i≤n), the value of elements in the sequence.

Output

For each test case, print a line containing the test case number (beginning with 1) and the number of positive sequences.

Sample Input

2
3
1 1 -1
8
1 1 1 -1 1 1 1 -1

Sample Output

Case 1: 1
Case 2: 4

思路：可以建立一个长度为2*n的数组，每次查询它的前ai项之和是不是负数，如果是负数，

用vis标记，可以保证后面扫的时候，扫到这个地方就可以停下了

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<cmath>

const int maxn=5e5+5;
typedef long long ll;
using namespace std;
ll a[2*maxn];
int vis[maxn];
int main()
{
    int t,cnt=1,i,n,j;
    scanf("%d",&t);
    while(t--)
    {
        memset(a,0,sizeof(a));
        memset(vis,0,sizeof(vis));
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%I64d",&a[i]);
            a[i+n]=a[i];
        }
        for(i=2*n-1;i>=n;i--)
        {
            if(i<n) break;
            if(a[i]<=0)
            {
                ll sum=0;
                for(j=i;j>i-n;j--)
                {
                    sum+=a[j];
                    if(sum<=0)
                    {
                        if(j<n) vis[j]=1;
                        else vis[j-n]=1;
                    }
                    else break;
                }
                i=j;
            }
            if(i<n) break;
        }
        int ans=0;
        for(i=0;i<n;i++)
        {
            if(vis[i]==1) ans++;
        }
        
        printf("Case %d: %d
",cnt++,n-ans);
    }
    
    return 0;
}