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文本数字提取与求和的Java程序解析

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Java是一种广泛使用的面向对象的编程语言,它具有高度的可移植性、安全性和性能优势。Java被广泛应用于服务器端应用程序、大型系统开发、Android应用开发等领域。在处理文本文件和执行数值计算方面,Java提供了强大的API和工具集。在本案例中,我们将讨论如何使用Java来分析一个包含数字的文本文件,并执行特定的任务:查找并求和所有的数字。 任务描述中提到的“查找数字并添加”,要求编写一个程序来处理一段文本,找到所有的数字并将它们的和计算出来。这个问题不仅仅是一个简单的数字求和问题,更是一个文本解析和字符串处理的问题。在这个过程中,我们需要识别出文本中的数字形式(如“一千亿”、“十万光年”、“三万光年”等),然后将这些数字从文字形式转换为数值形式,并计算它们的总和。 在进行编码实现之前,需要对文本中可能出现的数字进行分类。文本中的数字形式包括整数、小数、以及具有特定量词的数字(例如“一百万”、“一万六千”等)。在转换这些数字时,我们可能需要创建一个字典或映射来将特定的量词映射到它们的数值(例如“十”对应10,“百”对应100,“千”对应1000,“万”对应10000等),然后将这些数值与数字的主体部分结合,以正确计算总和。 举个例子,文本中的“一百万”首先需要识别“百”和“万”对应的数值,然后将这些数值与主体数字“一”结合起来,最后将得到的数值1000000加到总和中。同理,对于“十万光年”,我们需要计算出100000,并将其加到总和中。 在实现这个程序时,有以下步骤是关键: 1. 读取文本文件:Java提供了多种读取文件的方法,比如使用`BufferedReader`或者`Files`类从文件中逐行或逐字符读取数据。 2. 文本解析:需要编写代码来解析文本,识别数字和相关量词。这可能需要编写一个正则表达式来匹配文本中的数字模式。 3. 数字转换:将匹配到的文字形式的数字转换成数值。这需要处理单位的换算,如将“一百万”转换为1000000。 4. 数值计算:使用循环和条件判断来累加转换后的数值。 5. 输出结果:将最终的总和输出到控制台或写入到文件。 考虑到这是一个通用解决方案,程序需要能够处理任何包含1万亿以下数字的文本,这意味着程序要有良好的健壮性和错误处理机制,以确保它能够应对各种格式的输入。 Java的异常处理机制,比如`try-catch`语句,可以帮助开发者处理在文件读取和数字转换过程中可能出现的错误。同时,还需要对最终计算结果进行格式化输出,确保结果的准确性和可读性。 考虑到这个文件名“Find-Numbers-and-Add-master”,它可能是GitHub上的一个项目仓库名称。该项目的目标就是实现上述的程序功能,并且很可能是用Java编写的。如果你需要查看该项目的源代码或进一步的详细信息,可以通过访问这个项目仓库来进行探索和学习。在该项目中,你可能会找到处理文本、解析数字、执行计算等相关的Java类和方法的实现细节。

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{"id": "cmpl-2683988a-7a7b-11f0-b2b4-1a85109d8f07", "choices": [{"finish_reason": "stop", "index": 0, "message": {"content": "<think>\nOkay, so I need to find the dot product of these two vectors: $\\begin{pmatrix} -2 \\\\ 0 \\\\ 7 \\end{pmatrix}$ and $\\begin{pmatrix} 3 \\\\ 4 \\\\ -5 \\end{pmatrix}$. Hmm, let me remember how the dot product works. From what I recall, the dot product involves multiplying corresponding components of the vectors and then adding them all up. Let me verify that.\n\nSo, if I have two vectors, say $\\mathbf{a} = \\begin{pmatrix} a_1 \\\\ a_2 \\\\ a_3 \\end{pmatrix}$ and $\\mathbf{b} = \\begin{pmatrix} b_1 \\\\ b_2 \\\\ b_3 \\end{pmatrix}$, their dot product $\\mathbf{a} \\cdot \\mathbf{b}$ is calculated as $a_1b_1 + a_2b_2 + a_3b_3$. Yeah, that sounds right. It's like multiplying each pair of components in the same position and then summing all those products. Got it.\n\nAlright, applying this to the given vectors. Let me label them for clarity. Let vector $\\mathbf{u} = \\begin{pmatrix} -2 \\\\ 0 \\\\ 7 \\end{pmatrix}$ and vector $\\mathbf{v} = \\begin{pmatrix} 3 \\\\ 4 \\\\ -5 \\end{pmatrix}$. So, following the formula, I need to compute $u_1v_1 + u_2v_2 + u_3v_3$.\n\nLet me write that out step by step. First, multiply the first components: $-2$ times $3$. Let me calculate that: $-2 \\times 3 = -6$. Okay, that's straightforward.\n\nNext, multiply the second components: $0$ times $4$. Hmm, $0 \\times 4 = 0$. That's easy enough. The zero will just drop out of the sum.\n\nThen, multiply the third components: $7$ times $-5$. Let me do that: $7 \\times -5 = -35$. Right, multiplying a positive by a negative gives a negative, so that's -35.\n\nNow, I need to add up all those results: $-6 + 0 + (-35)$. Let me compute that. Starting with -6 plus 0 is still -6. Then, adding -35 to that would be like subtracting 35 from -6. So, $-6 - 35$. Let me think... that should be $-41$. Wait, is that right? Let me check again.\n\nYes, if you have a debt of $6 and then you add a debt of $35, your total debt is $41. So, in mathematical terms, that's -41. Okay, so the total sum is -41. Therefore, the dot product of these two vectors is -41.\n\nWait, let me just make sure I didn't mix up any signs. The first term was -2*3=-6, second term 0*4=0, third term 7*(-5)=-35. Adding them up: -6 + 0 is -6, then -6 + (-35) is indeed -41. Yeah, that seems correct.\n\nIs there another way to check this? Maybe by using the formula for the dot product in terms of magnitudes and the cosine of the angle between them? Although I don't know the angle here, so maybe that's not helpful. But just to recall, the dot product can also be expressed as $|\\mathbf{u}||\\mathbf{v}|\\cos{\\theta}$, where $\\theta$ is the angle between them. But since I don't have the angle, I can't use that here for verification. So maybe I should stick with the component method.\n\nAlternatively, maybe I can recalculate the component-wise products again. First component: -2*3. Let me do 2*3=6, and since one is negative, it's -6. Second component: 0*4. 0 times anything is 0. Third component: 7*(-5). 7*5 is 35, and since one is negative, it's -35. Adding them up: -6 + 0 + (-35) = -41. Yep, same result. So I think that's solid.\n\nWait, but let me check once more. Maybe I made a mistake in arithmetic. Let me write it numerically:\n\nFirst term: (-2)(3) = -6\n\nSecond term: (0)(4) = 0\n\nThird term: (7)(-5) = -35\n\nTotal: (-6) + 0 + (-35) = (-6) + (-35) = -(6 + 35) = -41. Yep, that's correct. 6 + 35 is 41, so negative of that is -41. Perfect.\n\nSo, I think I can be confident that the dot product is -41. Let me just recall if there are any common mistakes people make here. For example, sometimes people might confuse the dot product with the cross product, but cross product gives a vector, whereas dot product is a scalar. Since the question specifically asks for the dot product, which is a scalar, and we have a scalar here, -41, that seems okay.\n\nAnother common mistake might be mixing up the signs. For instance, if someone thought that multiplying two negatives gives a negative, but no, two negatives give a positive. However, in our case, only the first and third components had negative numbers. Let me check again:\n\nFirst component: -2 (negative) and 3 (positive): negative times positive is negative. Correct.\n\nSecond component: 0 and 4: 0, no issues.\n\nThird component: 7 (positive) and -5 (negative): positive times negative is negative. Correct. So the signs are handled properly.\n\nAnother possible mistake could be miscalculating the products. Let me check each multiplication again:\n\n-2 * 3: 2*3 is 6, with a negative sign: -6. Correct.\n\n0 * 4: 0. Correct.\n\n7 * -5: 7*5 is 35, with a negative sign: -35. Correct.\n\nSo all multiplications are correct. Then adding them up: -6 + 0 + (-35). Adding two negative numbers: -6 -35. Which is - (6 + 35) = -41. Correct.\n\nTherefore, I think my answer is correct. The dot product is indeed -41.\n\n**Final Answer**\nThe dot product is \\boxed{-41}.\n</think>\n\nTo compute the **dot product** of the two vectors:\n\n$$\n\\begin{pmatrix} -2 \\\\ 0 \\\\ 7 \\end{pmatrix} \\quad \\text{and} \\quad \\begin{pmatrix} 3 \\\\ 4 \\\\ -5 \\end{pmatrix},\n$$\n\nwe use the standard formula for the dot product of two 3-dimensional vectors:\n\n$$\n\\mathbf{u} \\cdot \\mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3\n$$\n\n---\n\n### Step-by-step Calculation:\n\nLet’s label the vectors for clarity:\n\n- $\\mathbf{u} = \\begin{pmatrix} -2 \\\\ 0 \\\\ 7 \\end{pmatrix}$\n- $\\mathbf{v} = \\begin{pmatrix} 3 \\\\ 4 \\\\ -5 \\end{pmatrix}$\n\nNow compute the product of corresponding components:\n\n1. $u_1 \\cdot v_1 = -2 \\cdot 3 = -6$\n2. $u_2 \\cdot v_2 = 0 \\cdot 4 = 0$\n3. $u_3 \\cdot v_3 = 7 \\cdot (-5) = -35$\n\n---\n\n### Sum the Results:\n\n$$\n-6 + 0 + (-35) = -6 - 35 = -41\n$$\n\n---\n\n### Final Answer:\n\n$$\n\\boxed{-41}\n$$", "role": "assistant", "reasoning_content": ""}, "review": {"gold": "-41", "pred": "-41", "result": true}}], "created": 1755332979, "model": "Qwen3-32B", "object": "chat.completion", "usage": {"completion_tokens": 1772, "prompt_tokens": 71, "total_tokens": 1843}, "model_spec": {"api_url": "http://0.0.0.0:9009/v1/chat/completions", "model_id": "Qwen3-32B", "api_key": "EMPTY"}, "answer_id": "answer-8c0749871bdda6509f7d96aead28ce76", "subset_name": "Level 1", "raw_input": {"problem": "Find the dot product of $\\begin{pmatrix} -2 \\\\ 0 \\\\ 7 \\end{pmatrix}$ and $\\begin{pmatrix} 3 \\\\ 4 \\\\ -5 \\end{pmatrix}.$", "solution": "The dot product of $\\begin{pmatrix} -2 \\\\ 0 \\\\ 7 \\end{pmatrix}$ and $\\begin{pmatrix} 3 \\\\ 4 \\\\ -5 \\end{pmatrix}$ is\n\\[(-2)(3) + (0)(4) + (7)(-5) = \\boxed{-41}.\\]", "answer": "-41", "subject": "Precalculus", "level": 1, "unique_id": "test/precalculus/1289.json"}, "origin_prompt": {"data": ["Find the dot product of $\\begin{pmatrix} -2 \\\\ 0 \\\\ 7 \\end{pmatrix}$ and $\\begin{pmatrix} 3 \\\\ 4 \\\\ -5 \\end{pmatrix}.$\nPlease reason step by step, and put your final answer within \\boxed{}."], "index": 30, "system_prompt": null, "raw_input": {"problem": "Find the dot product of $\\begin{pmatrix} -2 \\\\ 0 \\\\ 7 \\end{pmatrix}$ and $\\begin{pmatrix} 3 \\\\ 4 \\\\ -5 \\end{pmatrix}.$", "solution": "The dot product of $\\begin{pmatrix} -2 \\\\ 0 \\\\ 7 \\end{pmatrix}$ and $\\begin{pmatrix} 3 \\\\ 4 \\\\ -5 \\end{pmatrix}$ is\n\\[(-2)(3) + (0)(4) + (7)(-5) = \\boxed{-41}.\\]", "answer": "-41", "subject": "Precalculus", "level": 1, "unique_id": "test/precalculus/1289.json"}}, "reviewed": true, "review_id": "review-609e2af201a0d1b26312788b5e4bad31", "reviewer_spec": {"metric": ["AveragePass@1"], "reviewer": ["Evaluator"], "revision": ["default"]}, "review_time": 1755333711.497056} 分析其中关键信息

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二、实验目的 通过实验掌握下列知识: 1) 加深对顺序结构的理解。 2)掌握顺序程序的设计方法。 3)学习子程序的定义和调用方法。 4)掌握子程序、子程序的嵌套、递归子程序的结构。 5) 掌握子程序设计、编制及调试。 6) 熟练掌握DEBUG的常用命令,学会用DEBUG调试程序。 三、实验要求 1) 做好实验预习,完成各种软件安装与操作。 2) 按照实验内容与步骤的要求,独立思考,认真仔细地完成实验。 3) 截取各步骤截图,写出实验报告。 四、实验内容 1)计算1+2+3+…+10,将结果显示在屏幕上。 2)利用01H号功能调用输入10个一位数字,并将其由ASCII码转换为二进制数,依此保存到变量BUF的10个字节中,变量BUF的形式为BUF DB 10 DUP(?)。编程求出这10个数中的最大数和最小数,将最大数存入MAX单元、最小数存入MIN单元,并将其在屏幕上显示出来。 参考代码: 实验1 code segment assume CS:code start: ;程序入口 ;计算1+2+3+…+10,累加和=AL mov AL,0 mov CX,10 ;循环次数 next: ;循环入口 add AL,CL loop next ;将累加和AL显示在屏幕上,AL=55----->'5' '5' ;1.求十位数字 mov AH,0 mov BL,10 ;除数 div BL ;商=AL,余数=AH add AL,30H ;把AL中的数转为相应字符,如5-->‘5’ ;2.求个位数字 add AH,30H ;个位数字字符 ;输出十位数字与个位数字 mov DL,AL push AX mov AH,2 int 21H pop AX mov DL,AH mov AH,2 int 21H ;程序退出 mov AH,4cH int 21H code ends end start

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方2郭
2025.05.22
实用性高,适用于处理大量文本数据中的数字信息。
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网络小精灵
2025.04.29
文档描述生动,用宇宙的宏大比喻程序的潜力,引人入胜。🍘
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药罐子也有未来
2025.04.09
代码简洁,使用Java编写,易于理解和应用。
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马克love
2025.03.28
程序设计巧妙,能够将文本中的数字提取并进行求和。
缪之初
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