作业七-ID-1050-Problem B: Sequence Problem : Array Practice

本文介绍了一个涉及数组操作和条件输出的C语言程序案例。主要内容包括使用两个数组来存储交替输入的数据,并根据数组长度的不同实现相应的输出逻辑。文章还强调了初始化数组及处理特殊情况的重要性。

题目:


*****************************************************************************************************************************************************************************************解题思路:

首先,这个题要定义两个数组,用来储存两次输入,奇数时和偶数时。

在奇数次输入时,只进行输入。

在偶数次输入时,进行输入后要判断两数组长度进行输出。

*注意在总次数为奇数时要处理最后一次的输出。

*注意在输出时讨论两组数组都为空的情况。

*注意初始化数组。

*****************************************************************************************************************************************************************************************

给出代码:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
    int a[1010],b[1010],c[1010],add[1010];
    int m,n,i,j,k,x1,x2;
    k=1;
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    while(scanf("%d",&n)!=EOF)
     {
         if(k==1)
         {   if(n==0)
              {
                  x1=n;
 
              }
              else
              {
                  for(i=1;i<=n;i++)
             {   x1=n;
                 scanf("%d",&a[i]);
             }
              }
             k++;
         }
         else
         {
             if(k%2==0)
             {   x2=n;
                 k++;
                 if(x1==0&&x2==0)
                 {
                     printf("\n");
                     memset(a,0,sizeof(a));
                     memset(b,0,sizeof(b));
                 }
                 else{
                    for(i=1;i<=n;i++)
                 {
                     scanf("%d",&b[i]);
                 }
                   if(x1>=x2)
                   {
                       for(i=1;i<=x1;i++)
                       {
                           if(i==1)
                            printf("%d",a[i]+b[i]);
                           else
                            printf(" %d",a[i]+b[i]);
                       }
                   }
                   else
                   {
                       for(i=1;i<=x2;i++)
                       {
                           if(i==1)
                            printf("%d",a[i]+b[i]);
                           else
                            printf(" %d",a[i]+b[i]);
                       }
                   }
                   printf("\n");
 
 
             memset(a,0,sizeof(a));
             memset(b,0,sizeof(b));
             }
                 }
 
             else
             {
                if(n==0)
              {
                  x1=n;
 
              }
              else
              {
                  for(i=1;i<=n;i++)
             {   x1=n;
                 scanf("%d",&a[i]);
             }
              }
             k++;
             }
         }
 
     }
     if(k%2==0)
     {
         if(x1==0)
            printf("\n");
         else
         {
             for(i=1;i<=x1;i++)
             {
                 if(i==1)
                    printf("%d",a[i]);
                 else
                    printf(" %d",a[i]);
             }
             printf("\n");
         }
     }
     return 0;
}


\documentclass[12pt]{article} \usepackage{amsmath, amssymb} \usepackage{graphicx} \usepackage{geometry} \usepackage{setspace} \usepackage{caption} \usepackage{titlesec} % 页面设置 \geometry{a4paper, margin=1in} \onehalfspacing % 调整章节标题格式 \titleformat{\section}{\large\bfseries}{\thesection}{1em}{} \titleformat{\subsection}{\normalsize\bfseries}{\thesubsection}{1em}{} % 论文信息 \title{Sieve of Eratosthenes} \author{Zhang Hongwei} \date{December 2, 2025} \begin{document} \maketitle \begin{abstract} This paper describes the Sieve of Eratosthenes, an ancient algorithm for identifying all prime numbers up to a given limit $ n $. The method works by iteratively marking the multiples of each prime starting from 2. We outline its procedure, justify key optimizations, analyze time and space complexity, and compare it with modern variants. A flowchart is included to illustrate the execution process. \end{abstract} \section{Introduction} Finding all primes less than or equal to $ n $ is a basic problem in number theory. While checking individual numbers for primality can be done by trial division, generating many primes efficiently requires a different approach. The Sieve of Eratosthenes, attributed to the Greek mathematician Eratosthenes in the 3rd century BCE, provides a simple and effective solution. It avoids expensive divisibility tests by eliminating composite numbers through multiplication: once a number is identified as prime, all of its multiples are marked as non-prime. Given a positive integer $ n $, the algorithm produces all primes $ \leq n $. Its time complexity is $ O(n \log \log n) $, and it uses $ O(n) $ memory. This makes it practical for $ n $ up to several million on modern computers. \section{Basic Idea} A prime number has no divisors other than 1 and itself. The sieve exploits the fact that every composite number must have at least one prime factor not exceeding its square root. Starting with a list of integers from 2 to $ n $, we proceed as follows: \begin{itemize} \item Mark 2 as prime, then mark all multiples of 2 greater than $ 2^2 = 4 $ as composite. \item Move to the next unmarked number (3), mark it as prime, and eliminate multiples starting from $ 3^2 = 9 $. \item Repeat this process for each new prime $ p $ until $ p > \sqrt{n} $. \end{itemize} After completion, all unmarked numbers are prime. \subsection*{Why start from $ p^2 $?} Any multiple of $ p $ less than $ p^2 $, say $ k \cdot p $ where $ k < p $, would have already been marked when processing smaller primes. For example, $ 6 = 2 \times 3 $ is removed during the pass for 2. Thus, there's no need to revisit these values. \subsection*{Why stop at $ \sqrt{n} $?} If a number $ m \leq n $ is composite, it can be written as $ m = a \cdot b $, with $ 1 < a \leq b $. Then: \[ a^2 \leq a \cdot b = m \leq n \quad \Rightarrow \quad a \leq \sqrt{n}. \] So $ m $ must have a prime factor $ \leq \sqrt{n} $. Therefore, scanning beyond $ \sqrt{n} $ is unnecessary. \section{Implementation Steps} Consider $ n = 100 $. We use a boolean array \texttt{prime[0..100]}, initialized to \texttt{true}. Set \texttt{prime[0]} and \texttt{prime[1]} to \texttt{false}. \begin{enumerate} \item Start with $ p = 2 $. Since \texttt{prime[2]} is true, mark $ 4, 6, 8, \dots, 100 $ as false. \item Next, $ p = 3 $ is unmarked. Mark $ 9, 15, 21, \dots $ (odd multiples $ \geq 9 $). \item $ p = 4 $ is already marked; skip. \item $ p = 5 $ is prime. Mark $ 25, 35, 45, \dots $ \item $ p = 7 $: mark $ 49, 77, 91 $ \item $ p = 11 > \sqrt{100} $, so stop. \end{enumerate} All indices $ i \geq 2 $ where \texttt{prime[i] == true} are prime. \begin{figure}[h!] \centering \includegraphics[width=0.7\linewidth]{Flowchart.jpg} \caption{Flowchart of the Sieve of Eratosthenes algorithm} \label{fig:flowchart} \end{figure} Figure~\ref{fig:flowchart} shows the control flow: initialization, loop over $ p $ from 2 to $ \sqrt{n} $, and marking multiples starting at $ p^2 $. \section{Complexity Analysis} \subsection{Time Usage} For each prime $ p \leq \sqrt{n} $, we mark about $ n/p $ elements. Summing over such $ p $: \[ T(n) \approx n \sum_{\substack{p \leq \sqrt{n} \\ p\ \text{prime}}} \frac{1}{p}. \] It is known from number theory that the sum of reciprocals of primes up to $ x $ grows like $ \log \log x $. So: \[ \sum_{p \leq \sqrt{n}} \frac{1}{p} \sim \log \log \sqrt{n} = \log(\tfrac{1}{2}\log n) = \log \log n + \log \tfrac{1}{2} \approx \log \log n. \] Hence, total time is $ O(n \log \log n) $. \subsection{Memory Requirement} The algorithm requires one boolean value per integer from 0 to $ n $, leading to $ O(n) $ space usage. \section{Variants and Practical Considerations} \begin{table}[h!] \centering \caption{Common methods for generating primes} \label{tab:methods} \begin{tabular}{|l|c|c|l|} \hline Method & Time & Space & Remarks \\ \hline Trial division (single number) & $O(\sqrt{n})$ & $O(1)$ & Simple, slow for batches \\ Standard sieve & $O(n \log \log n)$ & $O(n)$ & Good for $ n \leq 10^7 $ \\ Segmented sieve & $O(n \log \log n)$ & $O(\sqrt{n})$ & Reduces memory usage \\ Linear sieve (Euler) & $O(n)$ & $O(n)$ & Faster in theory, more complex \\ \hline \end{tabular} \end{table} In practice, the standard sieve performs well due to good cache behavior and low constant factors. For very large $ n $, segmented versions divide the range into blocks processed separately. The linear sieve improves asymptotic time by ensuring each composite is crossed off exactly once using its smallest prime factor, but the overhead often negates benefits for moderate inputs. \section{Conclusion} The Sieve of Eratosthenes remains a fundamental tool in algorithm design. Its simplicity allows easy implementation and teaching, while its efficiency supports real-world applications in cryptography, number theory, and data processing. Although newer algorithms exist, the original sieve continues to be relevant—especially when clarity and reliability matter more than marginal speed gains. With minor improvements, it scales well within typical computational limits. \section{References} \begin{thebibliography}{9} \bibitem{knuth} Donald E. Knuth. \textit{The Art of Computer Programming, Volume 2: Seminumerical Algorithms}. 3rd Edition, Addison-Wesley, 1997. ISBN: 0-201-89684-2. (See Section 4.5.4 for discussion of prime number sieves.) \bibitem{hardy} G. H. Hardy and E. M. Wright. \textit{An Introduction to the Theory of Numbers}. 6th Edition, Oxford University Press, 2008. ISBN: 978-0-19-921986-5. (Chapter 1 discusses prime numbers and includes historical notes on Eratosthenes.) \bibitem{pomerance} Carl Pomerance. \newblock “A Tale of Two Sieves.” \newblock \textit{Notices of the American Mathematical Society}, vol.~43, no.~12, pp.~1473–1485, December 1996. Available online: \url{https://www.ams.org/journals/notices/199612/199612FullIssue.pdf#page=1473} \bibitem{crandall} Richard Crandall and Carl Pomerance. \textit{Prime Numbers: A Computational Perspective}. 2nd Edition, Springer, 2005. ISBN: 978-0-387-25282-7. (A detailed treatment of sieve methods including Eratosthenes and segmented variants.) \bibitem{eratosthenes-original} Thomas L. Heath (Ed.). \textit{Greek Mathematical Works, Volume II: From Aristarchus to Pappus}. Harvard University Press (Loeb Classical Library), 1941. ISBN: 978-0-674-99396-7. (Contains surviving fragments and references to Eratosthenes’ work in ancient sources.) \end{thebibliography} \end{document} 修改错误 ,并且增加字数在2000字左右
最新发布
12-03
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