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): 2622    Accepted Submission(s): 860

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

题解：给n个数，a0,a1,...an，求ai,ai+1,...an,a1,a2,...ai-1这样的排列种数，使得所有的前k（1<=k<=n)个的和都大于等于0；

求前缀和，加倍序列。

要满足前k个和都>=0,只需最小值>=0,所以用单调队列维护一个最小的前缀和sum[i],(i>=j-n+1)，这样就保证了sum[j]-sum[i]最大，所以区间【j-n+1，i]最小。

维护一个单调队列代表终止位置的最小值从小到大；

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN = 1e6 + 100;
int num[MAXN];
int sum[MAXN];
int q[MAXN];
int main(){
    int n;
    while(~scanf("%d", &n), n){
        sum[0] = 0;
        for(int i = 1; i <= n; i++){
            scanf("%d", num + i);
            sum[i] = sum[i - 1] + num[i];
        }
        for(int i = n + 1; i <= 2*n; i++)
            sum[i] = sum[i - 1] + num[i - n];
       // for(int i = 0; i <= 2*n; i++)
       //     printf("%d ", sum[i]);puts("");
        int head = 0, tail = -1, ans = 0;
        for(int i = 1; i <= 2 * n; i++){
            while(head <= tail && sum[i] < sum[q[tail]])tail--;
            q[++tail] = i;
          //  printf("i = %d %d %d
", i, sum[q[head]], sum[i - n]);
            if(i > n && sum[q[head]] - sum[i - n] >= 0)ans++;
            while(head <= tail && q[head] <= i - n)head++;
        }
        printf("%d
", ans);
    }
    return 0;
}