codeforces 1199A City Day 水题

http://codeforces.com/problemset/problem/1199/A
For years, the Day of city N was held in the most rainy day of summer. New mayor decided to break this tradition and select a not-so-rainy day for the celebration. The mayor knows the weather forecast for the n days of summer. On the i-th day, ai millimeters of rain will fall. All values ai are distinct.

The mayor knows that citizens will watch the weather x days before the celebration and y days after. Because of that, he says that a day d is not-so-rainy if ad is smaller than rain amounts at each of x days before day d and and each of y days after day d. In other words, ad<aj should hold for all d−x≤j<d and d<j≤d+y. Citizens only watch the weather during summer, so we only consider such j that 1≤j≤n.

Help mayor find the earliest not-so-rainy day of summer.

Input
The first line contains three integers n, x and y (1≤n≤100000, 0≤x,y≤7) — the number of days in summer, the number of days citizens watch the weather before the celebration and the number of days they do that after.

The second line contains n distinct integers a1, a2, …, an (1≤ai≤109), where ai denotes the rain amount on the i-th day.

Output
Print a single integer — the index of the earliest not-so-rainy day of summer. We can show that the answer always exists.

Examples
Input
10 2 2
10 9 6 7 8 3 2 1 4 5
Output
3
Input
10 2 3
10 9 6 7 8 3 2 1 4 5
Output
8
Input
5 5 5
100000 10000 1000 100 10
Output
5
Note
In the first example days 3 and 8 are not-so-rainy. The 3-rd day is earlier.

In the second example day 3 is not not-so-rainy, because 3+y=6 and a3>a6. Thus, day 8 is the answer. Note that 8+y=11, but we don’t consider day 11, because it is not summer.

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;

const int maxn=1e5+5;

int n,x,y;
int a[maxn];

int main()
{
	scanf("%d%d%d",&n,&x,&y);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(int i=1;i<=n;i++)
	{
		bool flag=1;
		for(int j=1;j<=x&&i-j>=1;j++)
		{
			if(a[i-j]<=a[i])
			{
				flag=0;
				break;
			}
		}
		if(!flag)
			continue;
		for(int j=1;j<=y&&i+j<=n;j++)
		{
			if(a[i+j]<=a[i])
			{
				flag=0;
				break;
			}
		}
		if(flag)
		{
			printf("%d\n",i);
			break;
		}
	}
	return 0;
}