Gym 100513F - Ilya Muromets

最新推荐文章于 2023-07-12 18:48:18 发布

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本文深入探讨了伊利亚穆罗梅茨如何在有限的两次挥剑中，最大化地削减戈洛尼奇龙头部的火属性威力，以求在史诗般的战斗中取得胜利。通过分析头部数量、每次挥剑可斩数量及每个头部的火属性强度，提出了一种高效的战斗策略。

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Ilya Muromets is alegendary bogatyr. Right now he is struggling against Zmej Gorynych, a dragonwith n headsnumbered from 1 to n from left to right.

Making one sweep of swordIlya Muromets can cut at most k contiguous heads ofZmej Gorynych. Thereafter heads collapse getting rid of empty space betweenheads. So in a moment before the second sweep all the heads form a contiguoussequence again.

As we all know, dragons canbreathe fire. And so does Zmej Gorynych. Each his head has a firepower. Thefirepower of the i-th head is f_i.

Ilya Muromets has time forat most two sword sweeps. The bogatyr wants to reduce dragon's firepower asmuch as possible. What is the maximum total firepower of heads which Ilya cancut with at most two sword sweeps?

Input

The first line contains apair of integer numbers n and k (1 ≤ n, k ≤ 2·10⁵) — thenumber of Gorynych's heads and the maximum number of heads Ilya can cut with asingle sword sweep. The second line contains the sequence of integer numbersf₁, f₂, ..., f_n (1 ≤ f_i ≤ 2000), where f_i is the firepower ofthe i-th head.

Output

Print the required maximumtotal head firepower that Ilya can cut.

Sample test(s)

input

8 2
1 3 3 1 2 3 11 1

output

input

4 100
10 20 30 40

output

100

思路：

首先我是想先求一边连续K个数最大和，再把剩下的合并，再求一遍最大和

接着发现其实可以同时求两段K个数最大和

程序：
#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <cstring>
#include <cstdio>
#include <stack>
#include <cstdlib>
#include <cmath>
using namespace std;

const int N = 200005;

int max(int a, int b)
{
	if (a > b)
		return a;
	else
		return b;
}

struct node
{
	int x, y;
};

int f[N * 2];
__int64 sum[N * 2];

int main()
{
	int n, k;
	scanf("%d %d", &n, &k);

	memset(f, 0, sizeof(f));
	memset(sum, 0, sizeof(sum));

	int i;
	for (i = 1; i <= n; i++)
	{
		scanf("%d", &f[i]);
		sum[i] = sum[i - 1] + f[i];
	}


	if (2 * k >= n)
	{
		printf("%I64d\n", sum[n]);
		return 0;
	}

	__int64 maxx = 0, maxy = 0;
	for (i = k; i <= n; i++)
	{
		if (sum[i] - sum[i - k] >= maxx)
		{
			maxx = sum[i] - sum[i - k];
		}
		maxy = max(maxy, maxx + sum[i + k] - sum[i]);
	}
	
	printf("%I64d\n", maxy);

	return 0;
}