F - Funny Positive Sequence（线性扫描）

最新推荐文章于 2021-06-17 10:31:37 发布

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本文介绍了一种算法解决方案，用于处理特定整数序列的正序移位问题。通过构建一个长度为2n的数组并标记所有可能的负前缀和位置，实现了对原始序列进行0至n-1次向前移位后的正序序列计数。

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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,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,…,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：就是输入一个a数组，然后对a进行向前移动i位的操作（0<=i<=n-1），求b数组的个数（bi是a1到ai的和）。

思路：直接按题目的意思做会超时，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，还有要用__int64,longlong不行。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdlib>
using namespace std;
typedef long long LL;
__int64 a[2*500500];
int vis[500500];
int main(void)
{
	int t,pt=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)
			{
				__int64 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 res=0;
		for(i=0;i<n;i++)
		{
			if(vis[i]==1) res++;
		}
		
		printf("Case %d: %d\n",pt++,n-res);
	}
	
	return 0;
}

参考文章：http://www.cnblogs.com/13224ACMer/p/5402326.html