cf Educational Codeforces Round 41 B. Lecture Sleep

本文主要是介绍cf Educational Codeforces Round 41 B. Lecture Sleep，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

原题：
B. Lecture Sleep
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Your friend Mishka and you attend a calculus lecture. Lecture lasts n minutes. Lecturer tells ai theorems during the i-th minute.

Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array t of Mishka’s behavior. If Mishka is asleep during the i-th minute of the lecture then ti will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — ai during the i-th minute. Otherwise he writes nothing.

You know some secret technique to keep Mishka awake for k minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and n - k + 1. If you use it on some minute i then Mishka will be awake during minutes j such that j∈[i,i+k-1] and will write down all the theorems lecturer tells.

You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.

Input
The first line of the input contains two integer numbers n and k (1 ≤ k ≤ n ≤ 105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.

The second line of the input contains n integer numbers a1, a2, … an (1 ≤ ai ≤ 104) — the number of theorems lecturer tells during the i-th minute.

The third line of the input contains n integer numbers t1, t2, … tn (0 ≤ ti ≤ 1) — type of Mishka’s behavior at the i-th minute of the lecture.

Output
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.

Example

input
6 3
1 3 5 2 5 4
1 1 0 1 0 0
output
16

Note
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16.

中文：

有个孩子听讲座，但是他总犯困，讲座持续n个小时。

给你两个行数据，每行有n个数，第一行代表每个小时讲座的内容量，第二行是一个0和1组成的行，0表示这孩子在睡觉，1表示他醒着。

这孩子在醒着的时候会记笔记，而且他有个本事，就是能让自己在连续的k个小时当中保持清醒，但是只能用一次，问你在整个n个小时的过程中，这孩子最多能记多少内容？

代码：

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;const int maxn = 100001;int n, k;
int a[maxn], sum[maxn];
bool mark[maxn];
int main()
{ios::sync_with_stdio(false);while (cin >> n >> k){memset(sum, 0, sizeof(sum));memset(mark, 0, sizeof(mark));for (int i = 1; i <= n; i++)cin >> a[i];for (int i = 1; i <= n; i++)cin >> mark[i];for (int i = 1; i <= n; i++){if (mark[i])sum[i] = sum[i - 1] + a[i];elsesum[i] = sum[i - 1];}int res = 0,ans=0;for (int i = 1; i <= k; i++)res += a[i];ans = res+sum[n]-sum[k];for (int i = k + 1; i <= n; i++){res += a[i];res -= a[i - k];ans = max(ans, res + sum[i - k] + sum[n] - sum[i]);}cout << ans << endl;}return 0;
}

解答：

滑动区间的问题，根据题目要求。

首先前k个小时都保持清醒时能记录的最多笔记是多少，以此作为初始值。

然后从k+1一直到n，枚举窗口的末端，判断区间中的“笔记值”全部记录，非区间中的笔记之按照是否为1，进行记录，更新最大值即可。

程序中使用sum[]来记录累加的所有“没睡着时的笔记值”，res记录滑动区间中所有的笔记值。

即
$ans = max(ans, res + sum[i - k] + sum[n] - sum[i]);$

sum[n]-sum[i]是区间右侧的“没睡着时的笔记值累加和”
sum[i - k] 是区间左侧的“没睡着时的笔记值累加和”

这篇关于cf Educational Codeforces Round 41 B. Lecture Sleep的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！

---