CodeForces 363B - Fence

最新推荐文章于 2024-07-28 21:06:44 发布

原创最新推荐文章于 2024-07-28 21:06:44 发布 · 1.1k 阅读

0 ·

CC 4.0 BY-SA版权

CodeForces 专栏收录该内容

16 篇文章

订阅专栏

本文介绍了一种算法，用于从一系列不同高度的围栏中找到连续的k块围栏，使其总高度最小。该问题源自将钢琴通过围栏缺口搬进家中的场景，通过计算连续部分围栏的高度总和来解决。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

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 itinto the house. In order to carry out his plan, he needs to take exactly k consecutive planks from the fence. Higher planksare harder to tear off the fence, so Polycarpus wants to find such k consecutive planks that the sum of their heightsis minimal possible.

Write the program that finds the indexes of k consecutive planks with minimal total height. Pay attention, the fenceis not around Polycarpus's home, it is in front of home (in other words, thefence 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 thefence and the width of the hole for the piano. The second line contains thesequence of integers h₁, h₂, ..., h_n (1 ≤ h_i ≤ 100), where h_i 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 suchj's, printany of them.

Sample test(s)

input

7 3
1 2 6 1 1 7 1

output

Note

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

思路：

找n个数中连续k个和最小

代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int main()
{
	int n, k;
	int a[150005];

	while (scanf("%d%d", &n, &k) != EOF)
	{
		int i, minn;
		memset(a, 0, sizeof(a));

		int flag = 1;

		for (i = 1; i<k; i++)
		{
			scanf("%d", &a[i]);
			a[i] += a[i - 1];
		}

		int temp;
		for (; i <= n; i++)
		{
			scanf("%d", &a[i]);
			a[i] += a[i - 1];
			temp = a[i] - a[i - k];
			if (temp<minn)
			{
				minn = temp;
				flag = i - k + 1;
			}
		}

		printf("%d\n", flag);
	}

	return 0;
}