HDU Non-negative Partial Sums （单调队列）

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

通过这个题才对单调队列有一点了解

题意：一个数列，每一次把第一个数放到尾部，判断这个数列对于任意的  i (1<=i<=n) 是否都满足 sum[i]>=0，如果满足ans++,求最大的ans

思路：先把串加倍，队列需要维护长度为n的序列中的最小值放在队首

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i < b; i++)
#define free(i,a,b) for(i = a; i > =b;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 2000005

int sum[N],a[N],n;
int que[N],tail,head;

void inque(int i)
{
    while(head<tail&&sum[i]<sum[que[tail-1]])
			tail--;
	que[tail++]=i;
}

void outque(int i)
{
     if(i-n>=que[head])
		head++;
}

int main()
{
	int i,j;
	while(~sf(n),n)
	{
		fre(i,1,n+1)
		 {
		 	sf(a[i]);
		    sum[i]=sum[i-1]+a[i];
		 }
        fre(i,n+1,n*2+1)
         sum[i]=sum[i-1]+a[i-n];

         tail=head=0;
         int ans=0;
         fre(i,1,n)
           inque(i);

		 fre(i,n,n*2)
		 {
		 	inque(i);
		 	outque(i);
		 	if(sum[que[head]]>=sum[i-n]) ans++;
		 }
       pf("%d\n",ans);
	}
   return 0;
}