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最新推荐文章于 2022-06-13 11:30:58 发布

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最新推荐文章于 2022-06-13 11:30:58 发布

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本文介绍了一个在线编程竞赛平台如何根据全球不同时间段的人数分布来选择最佳的比赛开始时间，确保最大程度地吸引参赛者。文章详细解释了通过计算各时段参与人数总和并找出峰值区间来确定最优开赛时间的算法。

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In distant future on Earth day lasts for n hours and that's why there are n timezones. Local times in adjacent timezones differ by one hour. For describing local time, hours numbers from 1 to n are used, i.e. there is no time "0 hours", instead of it "n hours" is used. When local time in the 1-st timezone is 1 hour, local time in the i-th timezone is i hours.

Some online programming contests platform wants to conduct a contest that lasts for an hour in such a way that its beginning coincides with beginning of some hour (in all time zones). The platform knows, that there are a_i people from i-th timezone who want to participate in the contest. Each person will participate if and only if the contest starts no earlier than s hours 00 minutes local time and ends not later than f hours 00 minutes local time. Values s and f are equal for all time zones. If the contest starts at f hours 00 minutes local time, the person won't participate in it.

Help platform select such an hour, that the number of people who will participate in the contest is maximum.

Input

The first line contains a single integer n (2 ≤ n ≤ 100 000) — the number of hours in day.

The second line contains n space-separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10 000), where a_i is the number of people in the i-th timezone who want to participate in the contest.

The third line contains two space-separated integers s and f (1 ≤ s < f ≤ n).

Output

Output a single integer — the time of the beginning of the contest (in the first timezone local time), such that the number of participants will be maximum possible. If there are many answers, output the smallest among them.

Example

Input

3
1 2 3
1 3

Output

Input

5
1 2 3 4 1
1 3

Output

Note

In the first example, it's optimal to start competition at 3 hours (in first timezone). In this case, it will be 1 hour in the second timezone and 2 hours in the third timezone. Only one person from the first timezone won't participate.

In second example only people from the third and the fourth timezones will participate.

题目的意思是，f-s的区间长度上选取人数最多的区间，该区间的开始设为s，反推第一区间的时间！

取模求解的方法很妙！

记得取模可能取到0，改为n！

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include<iomanip>
using namespace std;
typedef long long ll;
#define inf int(0x3f3f3f3f)
#define mod int(1e9+7)
#define pi acos(-1)
const int maxn = 1e5+10;
int n,s,f;
ll a[maxn],sum[maxn];

int main()
{
	while(~scanf("%d",&n))
	{
		sum[0]=0;
		for(int i=1;i<=n;i++)
		{
			scanf("%I64d",&a[i]);
			sum[i]=sum[i-1]+a[i];
		}
		scanf("%d%d",&s,&f);
		int num=f-s;
		int pos=0,ans;
		ll maxv=(ll)0;
		for(int i=num;i<=n+num;i++)
		{
			ll peo;
			pos=i-num;
			if(i<=n) peo=sum[i]-sum[pos];
			else peo=sum[n]-sum[pos]+sum[i-n];
			if(peo>maxv)
			{
				maxv=peo;
				ans=(n-(pos%n)+s)%n;
				if(ans==0)
				 ans=n;
			}
			else if(peo==maxv)
			{
				int res=(n-(pos%n)+s)%n;
				if(res==0) res=n;
				ans=min(res,ans);
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}