Codeforces Round #464 (Div. 2) C. Convenient For Everybody

C. Convenient For Everybody

time limit per test2 seconds
memory limit per test256 megabytes

Problem Description

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 ai 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, …, an (1 ≤ ai ≤ 10 000), where ai 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.

Examples

input
3
1 2 3
1 3
output
3
input
5
1 2 3 4 1
1 3
output
4

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.

---

解题心得：

1. 有n个时区，每个时区有一定的人数，给你一个区间[l，r)，要你选择一段连续的时区，长度为所给的区间长度，要求时区里面人数总和最多并且这个时区以l为当前的时间，输出第一个时区此时的时间。
2. 题目很绕，看了老半天，其实还是很简单的，首先要明白时间是循环的，先枚举得出每一段区间（长度为所给的区间长度）的人数总和，然后选出最大的那一段区间，逆推第一个时区的时间。

---

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5+100;
int num[maxn],n,l,r,len,sum[maxn],L,Max,ans;

void check(){
    int cnt = 2;
    Max = sum[1];
    l = 1,r = len;
    while(cnt <= n){
        r++;
        if(r > n)//时间是循环的
            r -= n;
        sum[cnt] = sum[cnt-1] - num[l] + num[r];//得到区间和
        l++;
        if(sum[cnt] > Max)
            Max = sum[cnt];
        cnt++;
    }
}

void get_ans(){
    ans = 1e18;
    for(int i=1;i<=n;i++){
        if(sum[i] == Max) {//找到最大的和并且第一个时区的时间最小
            int temp = L-i+1;
            if(temp <= 0)
                temp += n;
            if(temp < ans){
                ans = temp;
            }
        }
    }
}

int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&num[i]);
    scanf("%d%d",&l,&r);
    L = l;
    len = r-l;
    for(int i=1;i<=len;i++)
        sum[1] += num[i];
    check();
    get_ans();
    printf("%d",ans);
    return 0;
}