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最新推荐文章于 2020-07-25 15:23:54 发布

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#ACM #单调队列

探讨解决一个特定算法问题的方法：如何计算一个整数序列的所有循环移位中，满足其每个初始子序列和非负的移位数量。通过使用2*n倍数组处理环形数据并应用单调队列优化计算过程。

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Non-negative Partial Sums

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1323 Accepted Submission(s): 508

Problem Description

You are given a sequence of n numbers a ₀,..., a _n-1. A cyclic shift by k positions (0<=k<=n-1) results in the following sequence: a _k a _k+1,..., a _n-1, a ₀, a ₁,..., a _k-1. How many of the n cyclic shifts satisfy the condition that the sum of the fi rst i numbers is greater than or equal to zero for all i with 1<=i<=n?

Input

Each test case consists of two lines. The fi rst contains the number n (1<=n<=10 ⁶), the number of integers in the sequence. The second contains n integers a ₀,..., a _n-1 (-1000<=a _i<=1000) representing the sequence of numbers. The input will finish with a line containing 0.

Output

For each test case, print one line with the number of cyclic shifts of the given sequence which satisfy the condition stated above.

Sample Input

Sample Output

Source

SWERC 2011

开2*n倍数组处理这种环形数据，求sum数组，现在就是求所有的长度为n的序列的sumi的最小值与这个序列前一个sum的差是否大于0，因为区间是单调递增的，所以单调队列之

郁闷的是这道水题居然写了一晚上T_T，太久没写题了

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>

const int N = 1000100;

using namespace std;

int d[N+N],sum[N+N],q[N+N];

bool solve()
{
	int n,lim=0;
	scanf("%d",&n);
	if(n==0)return false;
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&d[i]);
		d[i+n]=d[i];
	}
	for(int i=1;i<=n+n;i++)
		sum[i]=sum[i-1]+d[i];
	
	int p1,p2,j,Ans=0;
	p1=1;p2=0;
	for(int i=1;i<n+n;i++)
	{
		//printf("i=%d  q[p1]=%d\n",i,q[p1]);
		while(p1<=p2 && sum[i]<sum[q[p2]])
			p2--;
		q[++p2]=i;
		if(i>=n)
		{
			while(q[p1]<=i-n)p1++;
			//printf("%d And %d\n",i-n,q[p1]);
			if(sum[q[p1]]>=sum[i-n])
				Ans++;
		}
	}
	printf("%d\n",Ans);
	return true;
}

int main()
{
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
	
	while(solve());
	
	return 0;
}