FZU 1914 Funny Positive Sequence

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本文介绍了一道算法题目，该题目要求对于给定的整数序列，通过循环左移生成多个新序列，并统计其中所有元素均为正数的序列数量。文章提供了完整的解题思路与C++代码实现。

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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,b2,…,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,…,an-1,an

A(1): a2,a3,…,an,a1

…

A(n-2): an-1,an,…,an-3,an-2

A(n-1): an,a1,…,an-2,an-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

题意: 给定一个序列，每个点可以做起点形成新序列，每个序列可以形成以头为起点所有长度的新序列Bi，如果所有Bi都大于0，那么这个序列是可以的，求一共几个这样的序列。

思路：逆向思维，从后往前推，如果当前位置<=0，那么则这个位置为起点是不行的，往前推，加到一个直到>0的sum，中间都是不符合的。 要注意一个点，就是如果加到起点，sum还是<=0的，那么要在从最后枚举一遍，因为序列是循环的。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
double a[500010];int vis[500020];
int main()
{
    int t;cin>>t;
    for(int kase=1;kase<=t;kase++)
    {
        int n;cin>>n;
        for(int i=0;i<n;i++){
            scanf("%lf",&a[i]);
        }
        int flag=1;double temp=0;int sum=n;
        memset(vis,0,sizeof(vis));
        for(int i=n-1;i>=0;i--){
            if(a[i]<=0&&flag)flag=0;
            if(!flag){
                temp+=a[i];
                if(temp<=0)vis[i]=1,sum--;
                else {flag=1;temp=0;}
            }
        }
        if(temp<=0)
        {
            for(int i=n-1;i>=0;i--)
            {
                temp+=a[i];
                if(temp<=0&&!vis[i]){
                    vis[i]=1;
                    sum--;
                }else if(temp>0)break;
            }
        }
        printf("Case %d: %d\n",kase,sum);
    }
}