hdu 4193 Non-negative Partial Sums 单调队列。

Non-negative Partial Sums

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1420    Accepted Submission(s): 544

Problem Description

You are given a sequence of n numbers a₀,..., 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, a₀, a₁,..., 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<=10⁶), the number of integers in the sequence. The second contains n integers a₀,..., 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

3

2 2 1

3

-1 1 1

1

-1

0

 

Sample Output

3

2

0

 

Source

SWERC 2011

 

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/*
题意：
刚刚又一道hdu的题目；
题意：10^6个数字，有10^6种形式。
如  a b c d ,  b c d a ,  c d a b,  d a b c;
统计在所以情况中如果满足任意前i数和都>=0 的个数。
想了一下，思路有了。可以这样子。
任意前i数和都满足>=0，那么最小的是不是也就满足了呢?
肯定的。所以。题目可以转换为
以i为开头，长度为n的前提下，求最小值。
这样的话，只有q[head].sum - s[ i - n];

 */

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
using namespace std;

int a[1000002];
int s[2000002];
typedef struct
{
    int num;
    int sum;
}Queue;
Queue tmp,q[2000004];

int main()
{
    int n,i,len,Num,tom;
    int head,tail;
    while(scanf("%d",&n)>0)
    {
        if(n==0)break;
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
            len=n*2;
        for(i=n+1;i<=len;i++)
            a[i]=a[i-n];
        for(s[0]=0,i=1;i<=len;i++)
            s[i]=s[i-1]+a[i];
        head=0;tail= -1; Num=1; tom=0;
        for(i=1;i<=len;i++)
        {
            tmp.num=++Num;
            tmp.sum=s[i];
            while( head<=tail && q[tail].sum>tmp.sum ) tail --;
            q[++tail]=tmp;
            if( i>n )
            {
                while( head<=tail && q[head].num+n<=i ) head++;
                if( q[head].sum-s[i-n]>=0) tom++;
            }
        }
        printf("%d
",tom);
    }
    return 0;
}