codeforces 363B Fence

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针对Polycarpus的问题，即如何从家中前方的围栏中找出连续的k块木板，使得这些木板的高度之和最小，本文提供了一个有效的算法解决方案。通过分析输入数据并运用滑动窗口技术，该方案能够快速定位到最佳起始位置。

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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 hi meters, distinct planks can have distinct heights.

$\text{[math]}$ 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 k consecutive 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 ≤ 1.5·105, 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
   
3

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.

暴力

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int a[150005];
    int temp,pre=0;
    int i,n,k;
    scanf("%d%d",&n,&k);
    for(i=0;i<n;i++)
    {
        scanf("%d",&temp);
        a[i]=temp+pre;
        pre=a[i];
    }
    int mi=a[k-1],seat=0;
    for(i=0;i<n-k;i++)
    {
        int w=a[i+k]-a[i];
        //printf("#%d %d\n",mi,w);
        if(mi>w)
        {
            mi=w;
            seat=i+1;
        }
        //printf("#%d %d\n",mi,w);
    }
    printf("%d\n",seat+1);
    return 0;
}