合肥生活安徽新聞合肥交通合肥房產生活服務合肥教育合肥招聘合肥旅游文化藝術合肥美食合肥地圖合肥社保合肥醫院企業服務合肥法律

        代寫 CS 336、代做 java/c++設計程序
        代寫 CS 336、代做 java/c++設計程序

        時間:2024-11-10  來源:合肥網hfw.cc  作者:hfw.cc 我要糾錯



        CS 336: Algorithms Problem Set 5 Date: Thursday, October 31, 2024 Due: Thursday, November 7, 2024
        Submit your solution on Gradescope.
        Please, solve all problems on your own. Do not collaborate with other students.
        Problem 1. The page limit for Problem 1 is 2 pages.
        Similarly to HW2, you want to travel from city A to city B located on a straight line (A is
        located in position 0 and B is located in position M ≥ 0), and you can travel at most distance D ≥ 0 miles per day, and you can only move to the right. Similarly, you have hotels between A and B with locations a1, . . . , an, where you can stay for a night.
        You are a person who likes to optimize all aspects of your life. In particular, if you didn’t fully use all D miles per day, it causes you great distress. Namely, if on some day you traveled distance d miles (out of possible D miles), the amount of distress is 2D−d.
        You start at city A. Your goal is to reach city B while suffering the least total amount of distress. Example: Assume that D = 4 and city B is located in position 6. You have two hotels in locations
        2 and 3. The following routes have the following distress:
        • 0→2→6: 24−(2−0) +24−(6−2) =4+1=5
        • 0→2→3→6: 24−(2−0) +24−(3−2) +24−(6−3) =4+8+2=14 • 0→2→6: 24−(3−0) +24−(6−3) =2+2=4
        The last route is optimal.
        Please do the following:
        • Formulate the subproblem. Please state it as precisely as possible. • Design a dynamic programming algorithm for solving this problem:
        – State the base case.
        – State the recurrence relation.
        – Explain why the recurrence relation is correct (from your explanation, one should un- derstand how to get your the recurrence relation).
        – Please provide the pseudocode. Please use the bottom-up approach.
        – Explain:
        ∗ What is the running time of your algorithm (all arithmetic operations take constant time).
        ∗ How to recover the maximum reward.
        ∗ How to recover the optimal route. You don’t need to write a pseudocode.
        ∗ How your algorithm correctly handles the case when an optimal solution doesn’t
        exist.
         1

        Problem 2. There is a new series in your streaming platform, Panopto. The series contains n episodes in total. Episodes need to be watched in order; that is, you cannot watch episode j before episode i if i < j. Since you’re busy, you decide to skip some subset of episodes (potentially empty). Your goal is to minimize the total amount of energy needed for this series, computed as follows:
        • You figure out that if you skip episode i, you would have to spend pi energy at the end of the year to figure out the missed content.
        • In addition, each episode has excitement value ei. You don’t want to dramatically change your emotions as well. So, for any consecutive episode i and j you watch, you need to spend |ei − ej | energy to adjust your mood as well.
        For example, if there are 5 episodes:
        • If you decide to watch episodes 1, 3, and 4, you need to spend p2 +p5 +|e1 −e3|+|e3 −e4| units of energy.
        • If you only decide to watch episode 3, you need to spend p1 + p2 + p4 + p5 units of energy.
        • If you decide to watch none of the episodes, you need to spend p1 +p2 +p3 +p4 +p5 units of
        energy.
        Implement the following function, which returns the list of episodes you decided to watch in the sorted order (the episodes are **indexed). For example, if you decide to watch first, third, and fourth episodes, your function must return a vector with items 1,3,4, in exactly this order. The input arrays are e and p respectively. It is guaranteed that for all test cases, the optimal answer is unique.
            vector<int> Episodes(const vector<int>& excitement, const vector<int>& penalty)
        Time limit The instructions are similar to the previous programming assignments. Your program should pass each tests in no more than 1 second. You can assume that 1 ≤ n ≤ 104 and all numbers are between 1 and 109.



        請加QQ:99515681  郵箱:99515681@qq.com   WX:codinghelp

        掃一掃在手機打開當前頁
      1. 上一篇:代做CMPT 401、代寫 c++設計程序
      2. 下一篇:代寫 CP3405、代做 Python/C++語言編程
      3. 無相關信息
        合肥生活資訊

        合肥圖文信息
        出評 開團工具
        出評 開團工具
        挖掘機濾芯提升發動機性能
        挖掘機濾芯提升發動機性能
        戴納斯帝壁掛爐全國售后服務電話24小時官網400(全國服務熱線)
        戴納斯帝壁掛爐全國售后服務電話24小時官網
        菲斯曼壁掛爐全國統一400售后維修服務電話24小時服務熱線
        菲斯曼壁掛爐全國統一400售后維修服務電話2
        美的熱水器售后服務技術咨詢電話全國24小時客服熱線
        美的熱水器售后服務技術咨詢電話全國24小時
        海信羅馬假日洗衣機亮相AWE  復古美學與現代科技完美結合
        海信羅馬假日洗衣機亮相AWE 復古美學與現代
        合肥機場巴士4號線
        合肥機場巴士4號線
        合肥機場巴士3號線
        合肥機場巴士3號線
      4. 短信驗證碼 酒店vi設計 投資移民

        關于我們 | 打賞支持 | 廣告服務 | 聯系我們 | 網站地圖 | 免責聲明 | 幫助中心 | 友情鏈接 |

        Copyright © 2025 hfw.cc Inc. All Rights Reserved. 合肥網 版權所有
        ICP備06013414號-3 公安備 42010502001045

        精品乱码久久久久久久| 国产成人无码精品一区不卡| 精品女同一区二区三区在线| 久久精品国产久精国产思思| 亚洲欧美日韩一区二区三区在线| 久久亚洲精品成人AV| 日产国产精品亚洲系列| 国产精品久久自在自线观看| 国产成人精品亚洲精品| 337P日本欧洲亚洲大胆精品| 亚洲AV无码之日韩精品| 久久99国内精品自在现线| 亚洲综合国产精品第一页 | 亚洲欧美日韩自偷自拍| 91精品啪在线观看国产电影 | 日韩国产成人精品视频| 亚洲精品无码你懂的网站| 国产精品无码一区二区在线| 亚洲高清专区日韩精品 | 国产系列高清精品第一页| 日韩精品无码一区二区三区四区 | 日韩精品久久不卡中文字幕| 99aiav国产精品视频| 国产亚洲精品AA片在线观看不加载| 国产精品videossex国产高清| 亚洲国产精品成人精品软件 | 中日韩精品视频在线观看| 国产精品午夜爆乳美女| 69久久夜色精品国产69| 久久久国产乱子伦精品| 99精品国产在热久久| 午夜国产精品无套| 国产日韩在线视频免费播放| 无码日本精品XXXXXXXXX| 3d动漫精品一区视频在线观看| 国产人妖乱国产精品人妖| 亚洲国产欧美日韩精品一区二区三区| 国产精品女同一区二区| 二区久久国产乱子伦免费精品| 全国精品一区二区在线观看| 日韩一级片免费观看|