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[];
 int s[];
 typedef struct
 {
     int num;
     int sum;
 }Queue;
 Queue tmp,q[];

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