363B. Fence

sliding window, prefix sum, 1100, https://codeforces.com/problemset/problem/363/B

原题目给出的tag是brute force, dp，应该是 sliding window, prefix sum

There is a fence in front of Polycarpus's home. The fence consists of n planks of the same width which go one after another from left to right. The height of the i-th plank is h**i meters, distinct planks can have distinct heights.

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Fence for n = 7 and h = [1, 2, 6, 1, 1, 7, 1]

Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly k consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such kconsecutive planks that the sum of their heights is minimal possible.

Write the program that finds the indexes of k consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic).

Input

The first line of the input contains integers n and k (1 ≤ n ≤ 1.5·10^5, 1 ≤ k ≤ n) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers h1, h2, ..., hn (1 ≤ hi ≤ 100), where hi is the height of the i-th plank of the fence.

Output

Print such integer j that the sum of the heights of planks j, j + 1, ..., j + k - 1 is the minimum possible. If there are multiple such j's, print any of them.

Examples

input

7 3
1 2 6 1 1 7 1

output

Note

In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8.

题意：给出 n 个木板高度数组 h[0..n-1]，要找到长度为 k 的连续子段，使得这个子段的总高度最小，输出子段的起始下标（从 1 开始）。

滑动窗口思路：滑动窗口：先算出第一个窗口的和，之后每次加一个元素、减一个元素即可，O(n)。

python

n, k = map(int, input().split())
h = list(map(int, input().split()))

# 先算第一个窗口的和
window_sum = sum(h[:k])
min_sum = window_sum
min_index = 0  # 从 0 开始计

# 滑动窗口
for i in range(k, n):
    window_sum += h[i] - h[i-k]
    if window_sum < min_sum:
        min_sum = window_sum
        min_index = i - k + 1

print(min_index + 1)  # 转成从 1 开始

前缀和思路是：

构造前缀和数组 prefix，其中 prefix[i] = h[0]+h[1]+⋯+h[i−1] （注意 prefix[0] = 0）。

那么长度为 k 的子段 [i, i+k-1] 的和就是：sum(i,i+k−1) = prefix[i+k] − prefix[i]

只要枚举所有起点 i (0 ≤ i ≤ n-k)，找最小的和。

python

n, k = map(int, input().split())
h = list(map(int, input().split()))

# 构建前缀和
prefix = [0] * (n + 1)
for i in range(n):
    prefix[i+1] = prefix[i] + h[i]

# 枚举长度为 k 的区间
min_sum = float('inf')
min_index = 0
for i in range(n - k + 1):
    window_sum = prefix[i+k] - prefix[i]
    if window_sum < min_sum:
        min_sum = window_sum
        min_index = i

print(min_index + 1)  # 转成 1-based

363B. Fence ​

363B. Fence