see-371. Sum of Two Integers

本文介绍了一种使用位运算实现两个整数相加的方法。通过异或运算实现无进位加法,与运算获取进位,并通过循环迭代直至没有进位完成整个加法过程。

利用位运算求二个整数的和:原理异或运算是无进制加法,与运算可求得进位

class Solution {
public:
    int getSum(int a, int b) {
        int res;
        int flag=a&b;//进位
        while(flag)
        {
            res=(a^b); //xor
            flag=flag<<1;//进位左移
            a=res;
            b=flag;
            flag=a&b;//进位
            
        }
        return a^b;
    }
};

 

r-906 & Hatsune Miku - All I Can See Is You Define a sequence of integers a valid if and only if ∀1≤i≤n,0≤ai≤i . Define the weight f(a) of a valid sequence a of length n : Initially, a token is placed at each integer point in the closed segment [1,n] on the number axis. Perform n operations in sequence. During the i -th operation, if ai≠0 , remove a token in the closed segment [ai,i] that has not been removed; otherwise, do nothing. f(a) is the number of ways to remove tokens. Two ways are considered different if there exists a t such that the positions of the tokens removed by the two ways are different at the t -th operation. For example, f([0,2,1])=2 , because we can remove tokens on 2,1 or 2,3 in sequence. JT gives you two integers n,m and asks you to find the sum of the weights over all (n+1)! valid sequences of length n . Since the answer may be too large, print it modulo m . Input Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤1000 ). The description of the test cases follows. The only line of each test case contains two integers n and m (1≤n≤5000,108≤m≤1.01⋅109 ) — the length of valid sequences, and the modulus. It is guaranteed that the sum of n2 over all test cases does not exceed 2.5⋅107 . Output For each test case, output an integer — the sum of the weights over all (n+1)! valid sequences of length n , modulo m . Example InputCopy 6 1 1000000007 2 1000000007 3 1000000007 4 1000000007 5 1000000007 114 514191981 OutputCopy 2 7 37 273 2672 393775292 Note In the first test case, valid sequences are [0] and [1] , and the answer is f([0])+f([1])=1+1=2 . In the second test case, valid sequences are [0,0],[0,1],[0,2],[1,0],[1,1],[1,2] . The weight of [0,1] is 2 , and the others are 1 , so the answer is 5⋅1+1⋅2=7 . 會用到啥概念
07-07
\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
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12-03
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