国产人妻人伦精品_欧美一区二区三区图_亚洲欧洲久久_日韩美女av在线免费观看

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CSC3100 Data Structures

時間:2023-11-27  來源:合肥網hfw.cc  作者:hfw.cc 我要糾錯



A. Requirements Code (**%)
You can write your code in Java, Python, C, or C++. The time limit may vary among different languages, depending on the performance of the language. Your code must be a complete excutable program instead of only a function. We guarantee test data strictly compliance with the requirements in the description, and you do not need to deal with cases where the input data is invalid.
Libraries in this assignment:
• For C/C++, you can only include standard library.
• For Java, you can only import java.util.*
• For Python, you can only import standard library. In other words, you cannot import libraries such as numpy.
We provide an example problem to illustrate the information above better.
Report (10%)
You also need to write a report in pdf type to explain the following: • What are the possible solutions for the problem?
• How do you solve this problem?
• Why is your solution better than others?
Please note that the maximum number of pages allowed for your report is 5 pages.
Remember that the report is to illustrate your thinking process. Keep in mind that your report is supposed to show your ideas and thinking process. We expect clear and precise textual descriptions in your report, and we do not recommend that you over-format your report.
B. Example Problem: A + B Problem Description
Given 2 integers A and B, compute and print A + B
Input
Two integers in one line: A, and B
Output
One integer: A + B
Sample Input 1 Sample Output 1
123
Problem Scale & Subtasks
For 100% of the test cases, 0 ≤ A,B ≤ 106
    1

Solutions
Java
import java.util.*;
public class Example {
public static void main(String[] args) {
int a, b;
Scanner scanner = new Scanner(System.in); a = scanner.nextInt();
b = scanner.nextInt();
scanner.close();
System.out.println(a + b);
} }
Python
AB = input (). split ()
A, B = int(AB[0]), int(AB[1]) print(A + B)
C
#include <stdio.h>
int main(int argc, char *argv[]) {
int A, B; scanf("%d%d", &A, &B); printf("%d\n", A + B); return 0;
}
C++
#include <iostream >
int main(int argc, char *argv[]) {
int A, B;
std::cin>> A >> B;
std::cout<< A + B << std::endl; return 0;
}
C. Submission
After finishing this assignment, you are required to submit your code to the Online Judge System (OJ), and upload your .zip package of your code files and report to BlackBoard.
C.1 Online Judge
Once you have completed one problem, you can submit your code on the page on the Online Judge platform (oj.cuhk.edu.cn, campus only) to gain marks for the code part. You can submit your solution of one problem for no more than 80 times.
After you have submitted your program, OJ will test your program on all test cases and give you a grade. The grade of your latest submission will be regarded as the final grade of the corresponding problem. Each problem is tested on multiple test cases of different difficulty. You will get a part of the score even if your algorithm is not the best.
        2

Note: The program running time may vary on different machines. Please refer to the result of the online judge system. OJ will show the time and memory limits for different languages on the corresponding problem page.
If you have other questions about the online judge system, please refer to OJ wiki (campus network only). If this cannot help you, feel free to contact us.
C.2 BlackBoard
You are required to upload your source codes and report to the BlackBoard platform. You need to name your files according to the following rules and compress them into A1_<Student ID>.zip :
A1_<Student ID>.zip
|-- A1_P1_<Student ID>.java/py/c/cpp |-- A1_P2_<Student ID>.java/py/c/cpp |-- A1_Report_<Student ID>.pdf
For Java users, you don’t need to consider the consistency of class name and file name. For example, suppose your ID is 123456789, and your problem 1 is written in Python, problem 2 is
written in Java then the following contents should be included in your submitted A1_123456789.zip:
A1_123456789.zip
|-- A1_P1_123456789.py
|-- A1_P2_123456789.java |-- A1_Report_123456789.pdf
C.3 Late Submissions
Submissions after Nov 24 2023 23:59:00(UTC+8) would be considered as LATE.
The LATE submission page will open after deadline on OJ.
Submisson time = max{latest submisson time for every problem, BlackBoard submisson time} There will be penalties for late submission:
• 0–24 hours after deadline: final score = your score×0.8 • 24–72 hours after deadline: final score = your score×0.5 • 72+ hours after deadline: final score = your score×0
FAQs
Q: I cannot access to Online Judge.
A: First, please ensure that you are using the campus network. If you are not on campus, please use the university VPN. Second, please delete cookies and refresh browser or use other browser. If you still cannot access to Online Judge, try to visit it via the IP address 10.26.200.13.
Q: My program passes samples on my computer, but not get AC on OJ. A: Refer to OJ Wiki Q&A
Authors
If you have questions for the problems below, please contact: • Yige Jiang: 1210**233@link.cuhk.edu.cn
• Ruiying Liu: ruiyingliu@link.cuhk.edu.cn
3

CSC3100 Data Structures Fall 2023 Programming Assignment 3
Due: Nov 24 2023 23:59:00
Assignment Link: http://oj.cuhk.edu.cn/contest/csc310023falla3 Access Code: 9v7Dxqet
1 Node Distance(40% of this assignment) 1.1 Description
You are given a tree with n nodes, where each edge in the tree has a corresponding weight denoting the length of each edge. The nodes in the tree are colored either black or white. Your task is to calculate the sum of distances between every pair of black nodes in the tree. Let B = {b1, b2, ...} a set of black nodes, then the answer is formulated as:
|B|−1 |B|
Ans= 􏰀 􏰀 dist(bi,bj)
i=1 j =i+1
where |B| denotes the number of the black nodes in the tree, and dist(bi,bj) is the length of the simple
path from the i-th to j-th black node.
Write a program to calculate the sum of distances on the tree between every pair of black nodes Ans
in the given tree.
1.2 Input
The first line contains an integer n, representing the number of nodes in the tree.
The second line contains n space-separated integers {c1,c2,...,ci,...,cn} where ci is either 0 or 1.
ci = 1 indicates that the i-th node is black, and ci = 0 indicates that the i-th node is white.
The following n − 1 lines, {l1, l2, . . . , lp, . . . , ln−1}, denoting the structure of the tree follow, each line lp contains 2 integers qp and wp, denoting an edge of length wp between the p + **th node and the qp-th node.
1.3 Output
Output the sum of distances for every pair of black nodes in the tree.
Sample Input 1 Sample Output 1
5 18 01111
11
12
** 31
    4

This sample considers a tree with 5 nodes:
 The **st node is white, and 2-, 3-, 4-, 5-th nodes are black.
The length of edge: (2-nd, **st): 1, (3-rd, **st): 2, (4-th, 3-rd): 2, (5-th, 3-rd): 1. Ans = ((1 + 2) + (1 + 2 + 2) + (1 + 2 + 1)) + (2 + 1) + 2 + 1 = 18.
Sample Input 2 Sample Output 2
9 96 010111111
12
13
22 21 52 53 12 71
Three additional large-scale samples are included in the provided files, namely, A samplecase1.in/.ans, A samplecase2.in/.ans and A samplecase3.in/.ans.
Problem Scale & Subtasks
For100%ofthetestcases,1≤n≤105,1≤qp−1 <p,1≤wp ≤1000
     Test Case No. **4
5-7 8
9 10
Hint
Constraints n ≤ 100
n ≤ 1000 qp = p
qp = 1
No additional constraints
1
12
23
21
45
  It can be proven that the given structure is definitely an unrooted tree.
For C/C++ and Java users, an int type stores integers range from -2,1**,483,648 to 2,1**,483,6**. It may be too small for this problem. You need other data types, such as long long for C/C++ and long for Java. They store integers ranging from -9,223,372,036,854,775,808 to 9,223,372,036,854,775,807. Use scanf("%lld",&n) for C, cin>>n for C++ and n = scanner.nextLong() for Java to get the input n. And the other operations for long and long long are quite same as int.
For Python users, if there occurs a RecusrionError, see here.
5

2 Price Sequence (50% of this assignment) 2.1 Description
Mario bought n math books and he recorded their prices. The prices are all integers, and the price sequence is a = {a0 , a2 , ...ai , ..., an−1 } of length n (n ≤ 100000). Please help him to manage this price sequence. There are three types of operations:
• BUY x: buyanewbookwithpricex,thusxisaddedattheendofa.
• CLOSEST ADJ PRICE: output the minimum absolute difference between adjacent prices.
• CLOSEST PRICE: output the absolute difference between the two closest prices in the entire se- quence.
A total of m operations are performed (1 ≤ m ≤ 100000). Each operation is one of the three mentioned types. You need to write a program to perform given operations. For operations ”CLOSEST ADJ PRICE” and ”CLOSEST PRICE” you need to output the corresponding answers.
2.2 Input
The first line contains two integers n and m, representing the length of the original sequence and the number of operations.
The second line consists of n integers, representing the initial sequence a.
Following that are m lines, each containing one operation: either BUY x, CLOSEST ADJ PRICE, or
CLOSEST PRICE (without extra spaces or empty lines).
2.3 Output
For each CLOSEST ADJ PRICE and CLOSEST PRICE command, output one line as the answer.
Sample Input 1
34
719 CLOSEST_ADJ_PRICE BUY 2 CLOSEST_PRICE CLOSEST_ADJ_PRICE
Sample Input 2
6 12
30 50 39 25 12 19 BUY 4 CLOSEST_PRICE
BUY 14 CLOSEST_ADJ_PRICE CLOSEST_PRICE
BUY 0 CLOSEST_PRICE
BUY 30
BUY 12 CLOSEST_PRICE
BUY 20 CLOSEST_PRICE
Sample Output 1
6 1 6
Sample Output 2
5 7 2 2 0 0
        Two additional large-scale samples are included in the provided files, namely, B samplecase1.in/.ans and B samplecase2.in/.ans.
6

Problem Scale & Subtasks
For 100% of the test cases, 2 ≤ n, m ≤ 1 × 105, 0 ≤ ai, x ≤ 1012
 Test Case No. **4
5-6 7-9 10
Hint
Constraints
n ≤ 103,m ≤ 103
There is no CLOSEST PRICE operation
ai and x are uniformly distributed at random within the range [0,1012] No additional constraints
  For C/C++ and Java users, an int type stores integers range from -2,1**,483,648 to 2,1**,483,6**. It may be too small for this problem. You need other data types, such as long long for C/C++ and long for Java. They store integers ranging from -9,223,372,036,854,775,808 to 9,223,372,036,854,775,807. Use scanf("%lld",&n) for C, cin>>n for C++ and n = scanner.nextLong() for Java to get the input n. And the other operations for long and long long 
請加QQ:99515681 或郵箱:99515681@qq.com   WX:codehelp

掃一掃在手機打開當前頁
  • 上一篇:ECE1747H代做、代寫python,Java程序
  • 下一篇:CS 2210編程代寫、Java程序語言代做
  • 無相關信息
    合肥生活資訊

    合肥圖文信息
    流體仿真外包多少錢_專業CFD分析代做_友商科技CAE仿真
    流體仿真外包多少錢_專業CFD分析代做_友商科
    CAE仿真分析代做公司 CFD流體仿真服務 管路流場仿真外包
    CAE仿真分析代做公司 CFD流體仿真服務 管路
    流體CFD仿真分析_代做咨詢服務_Fluent 仿真技術服務
    流體CFD仿真分析_代做咨詢服務_Fluent 仿真
    結構仿真分析服務_CAE代做咨詢外包_剛強度疲勞振動
    結構仿真分析服務_CAE代做咨詢外包_剛強度疲
    流體cfd仿真分析服務 7類仿真分析代做服務40個行業
    流體cfd仿真分析服務 7類仿真分析代做服務4
    超全面的拼多多電商運營技巧,多多開團助手,多多出評軟件徽y1698861
    超全面的拼多多電商運營技巧,多多開團助手
    CAE有限元仿真分析團隊,2026仿真代做咨詢服務平臺
    CAE有限元仿真分析團隊,2026仿真代做咨詢服
    釘釘簽到打卡位置修改神器,2026怎么修改定位在范圍內
    釘釘簽到打卡位置修改神器,2026怎么修改定
  • 短信驗證碼 豆包網頁版入口 破天一劍 目錄網 排行網

    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

    国产人妻人伦精品_欧美一区二区三区图_亚洲欧洲久久_日韩美女av在线免费观看
    亚洲精品久久久久久一区二区| 青青在线视频一区二区三区| 色狠狠久久av五月综合| 国产一区二区视频免费在线观看| 久久av一区二区三区亚洲| 久久久久国产精品一区| 国内精品一区二区三区四区| 久久精品99无色码中文字幕| 日本最新高清不卡中文字幕| 91免费版网站入口| 一区二区精品视频| 国产免费一区视频观看免费| 国产精品成av人在线视午夜片 | 黄色一级视频在线播放| 国产成人亚洲综合无码| 色综合久久久久无码专区| 91精品国产91久久久久久| 中文字幕无码不卡免费视频| 99热在线播放| 电影午夜精品一区二区三区| 91精品国产色综合久久不卡98| 亚洲在线视频福利| av观看免费在线| 亚洲蜜桃av| 高清视频在线观看一区| 欧美精品国产精品日韩精品| 国产视频九色蝌蚪| 色综合视频网站| 成人国产精品av| 亚洲综合在线做性| 97人人模人人爽人人喊38tv| 亚洲天堂第一区| 国产精品99免视看9| 日本不卡一区二区三区在线观看| 久久成人福利视频| 欧美不卡在线播放| 欧美精品一区二区免费| 福利精品视频| 亚洲v日韩v综合v精品v| 久久综合毛片| 欧美乱偷一区二区三区在线| 久久综合伊人77777蜜臀| 狠狠色综合色区| 久久99热精品这里久久精品| 99久久无色码| 日本成人中文字幕在线| 精品国产自在精品国产浪潮| 狠狠干 狠狠操| 欧美激情第6页| 91精品国产综合久久香蕉的用户体验 | 国产精品久久久久久久久久久不卡 | 97国产精品久久| 日本欧美一级片| 国产精品美女午夜av| 国产日产欧美视频| 亚洲成熟丰满熟妇高潮xxxxx| 久久一区二区三区av| 日韩欧美视频网站| 国产精品久久激情| 7777免费精品视频| 欧洲亚洲一区二区| 色与欲影视天天看综合网| 国产精品99一区| 黄色一级一级片| 亚洲一区二区三区久久| 久久精品一区中文字幕| 国产乱肥老妇国产一区二 | 成人国产精品av| 日韩亚洲在线视频| 欧美成aaa人片在线观看蜜臀| www.日本少妇| 热久久美女精品天天吊色| 国产精品后入内射日本在线观看| 国产免费黄色av| 日韩久久一级片| 一本—道久久a久久精品蜜桃| www.美女亚洲精品| 69久久夜色精品国产69| 国产在线一区二| 日韩精品无码一区二区三区| 一区二区精品国产| 国产精品久久九九| 国产精品∨欧美精品v日韩精品| 精品人妻一区二区三区四区在线| 亚洲一区二区三区精品视频| 国产精品日韩一区二区三区| 久久这里只有精品18| 国产欧美精品一区二区三区介绍 | 亚洲不卡一卡2卡三卡4卡5卡精品| 国产精品丝袜久久久久久不卡| 官网99热精品| 欧美日韩另类丝袜其他| 亚洲一区精彩视频| 国产精品丝袜久久久久久高清 | 国产欧美精品日韩精品| 欧洲精品久久久| 午夜精品在线视频| 欧美激情视频网| 国产精品日韩一区二区免费视频| 国产成人精品免费视频大全最热| 国产精品影院在线观看| 国内免费精品永久在线视频| 日韩精品欧美专区| 亚洲精品久久区二区三区蜜桃臀| 美女av一区二区| 国产精品久久久久av福利动漫| 久久精品xxx| 91精品国产综合久久男男| 国产欧美一区二区视频| 蜜臀av性久久久久蜜臀av| 欧美中文字幕在线| 日韩免费在线视频| 肉大捧一出免费观看网站在线播放| 欧美精品www| 精品久久久久久乱码天堂| 国产精品欧美日韩| 久久久91精品| 日韩亚洲成人av在线| 国产黄视频在线| 国产精品 欧美在线| 久久艳妇乳肉豪妇荡乳av| 91九色单男在线观看| 99www免费人成精品| 成人国产在线看| 97精品国产97久久久久久 | 欧美国产视频在线观看| 欧美在线精品免播放器视频| 奇米影视首页 狠狠色丁香婷婷久久综合 | 欧美 日韩 国产在线| 欧美午夜精品久久久久免费视| 日本欧美一二三区| 日韩视频第二页| 日韩欧美视频第二区| 日韩精品一区二区三区色欲av | 天堂资源在线亚洲资源| 亚洲精品欧洲精品| 亚洲xxxx做受欧美| 日韩av在线第一页| 日本精品va在线观看| 欧美中文字幕在线观看视频| 欧美高清一区二区| 国产中文字幕免费观看| 国产裸体舞一区二区三区| 高清一区二区三区四区五区| 91精品国产乱码久久久久久蜜臀| 国产成人精品免费视频大全最热 | 欧美精品在线播放| 中文字幕在线观看一区二区三区| 亚洲一区亚洲二区| 日本一区免费看| 欧美日韩一道本| 国产三级精品网站| 久久久一本二本三本| 日韩中文字幕在线视频| 国产精品久久久久9999小说| 亚洲综合成人婷婷小说| 日韩精品免费播放| 国产综合香蕉五月婷在线| 国产精品永久在线| 久久综合九色欧美狠狠| 久久久久免费精品| 国产精品久久久久不卡| 中文字幕一区二区三区四区五区| 视频一区二区三区免费观看| 欧美亚洲色图视频| 国产无套内射久久久国产| 91精品视频免费观看| 日韩在线中文字| 欧美精品在线免费| 日韩av免费一区| 国内精品视频在线| 国产精品自拍小视频| 久久久久久久影院| 国产99久久精品一区二区永久免费| 日韩一区二区三区资源| 精品人伦一区二区三区| wwwwww欧美| 精品国产一区二区三区久久久狼 | 色综合天天狠天天透天天伊人| 色综合电影网| 国产一区二区视频在线观看| 久久一区二区三区欧美亚洲| 国产精品九九九| 婷婷久久青草热一区二区| 国内精品久久国产| 国产高清在线一区| 久久综合久久88| 日本成人中文字幕在线| 成人免费aaa| 国产精品啪啪啪视频| 色狠狠久久av五月综合| 国产亚洲精品网站| 精品国产一区二区三区在线观看 | 青青久久av北条麻妃黑人| 高清视频一区二区三区| 久久精品视频99| 亚洲精品免费av| 国产肉体ⅹxxx137大胆| 日韩视频在线免费| 日本一区二区三区在线播放|