A. City Day
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
题意：输入n天数，前a天，后b天，求存在最早一天的降雨量均小于前a天，后b天，输出该天数。
思路：只需要找出某个区间中的最小数即可，对数组进行遍历即可。
代码
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
int a[100000];
int main()
{
int n, x, y, i, j, flag = 0;
scanf("%d %d %d", &n, &x, &y);
for (i = 0; i < n; i++)
{
scanf("%d", &a[i]);
}
for (i = 0; i < n; i++)
{
flag = 0;
for (j = i - x; j <= i + y; j++)
{
if (j < 0 || j >= n)
{
continue;
}
if (a[i] > a[j])
{
flag= 1;
}
}
if (flag == 0)
{
printf("%d\n", i + 1);
break;
}
}
return 0;
}