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 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

3

2 2 1

3

-1 1 1

1

-1

0

 

 

Sample Output

3

2

0

 

 

Source

SWERC 2011

 

 

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 1 /*

 2 题意：

 3 刚刚又一道hdu的题目；

 4 题意：10^6个数字，有10^6种形式。

 5 如  a b c d ,  b c d a ,  c d a b,  d a b c;

 6 统计在所以情况中如果满足任意前i数和都>=0 的个数。

 7 想了一下，思路有了。可以这样子。

 8 任意前i数和都满足>=0，那么最小的是不是也就满足了呢?

 9 肯定的。所以。题目可以转换为

10 以i为开头，长度为n的前提下，求最小值。

11 这样的话，只有q[head].sum - s[ i - n];

12 

13  */

14 

15 #include<iostream>

16 #include<stdio.h>

17 #include<cstring>

18 #include<cstdlib>

19 using namespace std;

20 

21 int a[1000002];

22 int s[2000002];

23 typedef struct

24 {

25     int num;

26     int sum;

27 }Queue;

28 Queue tmp,q[2000004];

29 

30 int main()

31 {

32     int n,i,len,Num,tom;

33     int head,tail;

34     while(scanf("%d",&n)>0)

35     {

36         if(n==0)break;

37         for(i=1;i<=n;i++)

38             scanf("%d",&a[i]);

39             len=n*2;

40         for(i=n+1;i<=len;i++)

41             a[i]=a[i-n];

42         for(s[0]=0,i=1;i<=len;i++)

43             s[i]=s[i-1]+a[i];

44         head=0;tail= -1; Num=1; tom=0;

45         for(i=1;i<=len;i++)

46         {

47             tmp.num=++Num;

48             tmp.sum=s[i];

49             while( head<=tail && q[tail].sum>tmp.sum ) tail --;

50             q[++tail]=tmp;

51             if( i>n )

52             {

53                 while( head<=tail && q[head].num+n<=i ) head++;

54                 if( q[head].sum-s[i-n]>=0) tom++;

55             }

56         }

57         printf("%d\n",tom);

58     }

59     return 0;

60 }