Foreword: This project is a part of pair programming task. We implement an command-line based arithmometer by JavaScript in Node.js environment. ......

Project Elaboration

In this project, We are required to actualize a command line program for automatically generating four arithmetic problems for primary schools.

Now the project is still in progress, we have so far finished exercise judgement, and soon we will update more information.

Team members: Wu Tang Huang, Zhang Guo Jun

GitHub: https://github.com/m8705/Arithmometer

Detail Description

Program usage

Generate problem file:

node e.js -n 10 -r 10

Validate exercise:

node e.js -e exercisefile.txt -a answerfile.txt

Parameter and regulation

  1. Use -n parameter to control the number of generated questions ( 1 ~ 10000 ).
  2. Use -r parameter to control the range of numeric value (natural number, true fraction and true fraction denominator) in the title ( 1 ~ 100 ).
  3. The calculation process in the generated problem does not produce a negative number, that is, if there is a sub-expression in the arithmetic expression, such as e1 - e2, then e1 < e2
  4. If there is a subexpression e1 ÷ e2 in the generated exercise, the result should be true score.
  5. There are no more than 3 operators in each problem.
  6. The problem generated by the program running at one time can not be repeated, that is, any two problems can not be transformed into the same problem by the finite number of exchange + and * arithmetic expressions.The generated problem is stored in the Exercises.txt file under the current directory of the execution program.
  7. At the same time, the answers to all the questions are computed and stored in the Answers.txt file in the current directory of the execution program.
  8. The program should support the generation of ten thousand problems.
  9. Program support for a given question file and answer file, determine the right and wrong answers and statistics, statistics output to the file Grade.txt

Code Preview

Convert fraction

 function convert(str){//将任何数转成真分数(小数不换)

     //整数 2 = 2'1/1
//真分数 3/8
//假分数 5/3
//带分数 1'1/2 //console.log(str) if( str.indexOf("/") >= 0 ){//真分数或带分数 if( str.indexOf("'") >= 0 ){//带分数 first = str.split("'")[0];
second = str.split("'")[1]; up = second.split("/")[0];
down = second.split("/")[1]; if( ( up === down ) || ( down === "1" ) ){//带分数情况下,不可能存在分子分母相同或分母为1的情况
return "ERROR";
} str = ( (+first) * (+down) + (+up) ) + "/" + down; }
else{//真分数
;
} }
else{//整数 str = str + "/1"; } return str
//console.log(str);
}

Fraction calculation

 function calculate(num1,num2,operator){//根据数字和符号进行分数运算

     var n1 = [];
var n2 = []; var result; n1 = convert(num1).split( "/" ); // [ 0分子,1分母 ]
n2 = convert(num2).split( "/" ); // [ 0分子,1分母 ] switch(operator){
case "+":
result = (n1[0]*n2[1]+n2[0]*n1[1]) + "/" + (n1[1]*n2[1]);
break;
case "-":
result = (n1[0]*n2[1]-n2[0]*n1[1]) + "/" + (n1[1]*n2[1]);
break;
case "*":
result = (n1[0]*n2[0]) + "/" + (n1[1]*n2[1]);
break;
case "/":
result = (n1[0]*n2[1]) + "/" + (n1[1]*n2[0]);
break;
} //console.log(result);
return result; }

Produce symbol

 function produceSymbol(){//产生符号

     var symbol = Math.random();
var symbolNum; if( symbol <= 1/3 ){//生成一个符号
symbolNum = 1;
}
else if( symbol <= 2/3 ){//生成两个符号
symbolNum = 2;
}
else{//生成三个符号
symbolNum = 3;
} var symbolChoice = [];
var tmp;
for(var a = 0; a < symbolNum; a++){//用概率决定符号 tmp = Math.random();
if( tmp <= 1/4 ){
symbolChoice.push("+");
}
else if( tmp <= 2/4 ){
symbolChoice.push("-");
}
else if( tmp <= 3/4 ){
symbolChoice.push("*");
}
else{
symbolChoice.push("/");
} } return symbolChoice; }

Produce number

 function produceNumber(symbolNum, range){//产生数字

     var symbolChoice = produceSymbol();

     var numType;
var numChoice = [];
var up, down; for( var b = 0; b < symbolNum + 1; b++ ){//用概率决定数字 numType = Math.random(); if( numType <= 7 / 10 ){//生成整数 numChoice.push( Math.floor(Math.random()*range) + "" ); }
else{//生成分数或1(避免生成分子或分母为0) up = Math.ceil( Math.random() * range );//向上取整
down = Math.ceil( Math.random() * range );//向上取整 if( up === down ){//分子分母相同
numChoice.push("1");
continue;
} var tmp = Math.random();//是否产生带分数
if( tmp <= 1/4 ){//产生带分数 while(up <= down || (up%down === 0) ){//重新产生带分数 up = Math.ceil( Math.random() * range );//向上取整
down = Math.ceil( Math.random() * range );//向上取整 } numChoice.push(
(up - up%down)/down +
"'" +
(up%down / gcd(up%down,down)) +
"/" +
down / gcd(up%down,down)
); }
else{//产生分数 numChoice.push( up + "/" + down ); } } }
return numChoice;
}

Produce array

 function produceRightArray(n, range){//产生n组符合规定的数字和符号

     var rightArray = [];
var flag; for(var a = 0; a < n; a++){//循环n次 flag = ""; symbolChoice = produceSymbol();
numChoice = produceNumber(symbolChoice.length,range); for(var b = 0; b < symbolChoice.length; b++ ){//遍历检查每个符号 if( symbolChoice[b] === "*" || symbolChoice[b] === "/" ){ if(numChoice[b] === "0" || numChoice[b+1] === "0"){ flag = "err";
a--;
break; } } } //console.log(a + flag); if(flag !== "err"){
rightArray.push([
symbolChoice,numChoice
]);
} } //console.log(rightArray);
return rightArray; }

Core computation

 function produceExercise(n,range){//产生n个习题(题目+答案)

     var expression = [];
var tmp = "";//保存用于产生结果的算式
var tmp1 = "";//保存用于产生题目的算式 var rightArray = produceRightArray(n,range); for(var a = 0; a < n; a++ ){//遍历每个产生的结果数组,分别验算结果是否非负 tmp = "";
tmp1 = ""
tmp += "(" + convert(rightArray[a][1][0]) + ")" ;
tmp1 += rightArray[a][1][0]; for(var b = 0; b < rightArray[a][0].length; b++ ){//符号+数字
tmp += rightArray[a][0][b] + "(" + convert(rightArray[a][1][b+1]) + ")";
tmp1 += " " + rightArray[a][0][b] + " " + rightArray[a][1][b+1];
} while( eval(tmp) < 0 ){//不允许产生算式最终值小于0的情况 rightArray[a] = produceRightArray(1,range)[0]; tmp = "";
tmp1 = "";
tmp += convert(rightArray[a][1][0]);
tmp1 += rightArray[a][1][0]; for(var c = 0; c < rightArray[a][0].length; c++ ){//符号+数字
tmp += rightArray[a][0][c] + "(" + convert(rightArray[a][1][c+1]) + ")";
tmp1 += " " + rightArray[a][0][c] + " " + rightArray[a][1][c+1];
} }
//console.log(tmp);
expression.push(tmp1);
} //console.log(expression) //console.log(rightArray); //遍历符号列表,根据优先级(先乘除,后加减)对数进行运算,并更新运算结果(逐一替换)至数组 var tmpArray = rightArray;
var operator; var symIndex;
var numIndex1, numIndex2; var answer = []; for(var d = 0; d < n; d++){ for(var e = 0; e < tmpArray[d][0].length; e++){//先进行乘除运算 operator = tmpArray[d][0][e]; switch(operator){ case "*":
//console.log(tmpArray[d][1][e],tmpArray[d][1][e+1]); replaceNumber(tmpArray[d][1],tmpArray[d][1][e],calculate( convert(tmpArray[d][1][e]),convert(tmpArray[d][1][e+1]),operator ) );
removeOperator(tmpArray[d][0],"*");
e--;
break; case "/":
//console.log(tmpArray[d][1][e],tmpArray[d][1][e+1]); replaceNumber(tmpArray[d][1],tmpArray[d][1][e],calculate( convert(tmpArray[d][1][e]),convert(tmpArray[d][1][e+1]),operator ) );
removeOperator(tmpArray[d][0],"/");
e--;
break; } } //console.log(tmpArray) for(var f = 0; f < tmpArray[d][0].length; f++){//后进行加减运算 operator = tmpArray[d][0][f]; switch(operator){ case "+":
//console.log(tmpArray[d][1][f],tmpArray[d][1][f+1]); replaceNumber(tmpArray[d][1],tmpArray[d][1][f],calculate( convert(tmpArray[d][1][f]),convert(tmpArray[d][1][f+1]),operator ) );
removeOperator(tmpArray[d][0],"+");
f--;
break; case "-":
//console.log(tmpArray[d][1][f],tmpArray[d][1][f+1]); replaceNumber(tmpArray[d][1],tmpArray[d][1][f],calculate( convert(tmpArray[d][1][f]),convert(tmpArray[d][1][f+1]),operator ) );
removeOperator(tmpArray[d][0],"-");
f--;
break; } } answer.push( simplify(tmpArray[d][1][0]) );
} //console.log(answer); return [expression,answer]; }

PSP 2.1

PSP2.1

Personal Software Process Stages

预估耗时(分钟)

实际耗时(分钟)

Planning

计划

 120  60

· Estimate

· 估计这个任务需要多少时间

 15  15

Development

开发

 360  300

· Analysis

· 需求分析 (包括学习新技术)

 60  60

· Design Spec

· 生成设计文档

 30  30

· Design Review

· 设计复审 (和同事审核设计文档)

 30  30

· Coding Standard

· 代码规范 (为目前的开发制定合适的规范)

 15  15

· Design

· 具体设计

 60  60

· Coding

· 具体编码

 120  100

· Code Review

· 代码复审

 120  160

· Test

· 测试(自我测试,修改代码,提交修改)

 40  40

· Reporting

· 报告

 90  90

· Test Report

· 测试报告

 60  60

· Size Measurement

· 计算工作量

 30  30

· Postmortem & Process Improvement Plan

· 事后总结, 并提出过程改进计划

 30  30

合计

  1180  1080

Screenshot

Produce

Judge

Project Summary

[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]][([][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[+!+[]+[+[]]]+([][[]]+[])[+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[+!+[]]+([][[]]+[])[+[]]+([][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[+!+[]+[+[]]]+(!![]+[])[+!+[]]]((![]+[])[+!+[]]+(![]+[])[!+[]+!+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]+(!![]+[])[+[]]+(![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[!+[]+!+[]+[+[]]]+[!+[]+!+[]+!+[]+!+[]+!+[]+!+[]]+[!+[]+!+[]+!+[]+!+[]+!+[]+!+[]]+[!+[]+!+[]+!+[]+!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[!+[]+!+[]+[+[]]])()

Arithmometer: A Node.js implementation的更多相关文章

  1. Edge.js:让.NET和Node.js代码比翼齐飞

    通过Edge.js项目,你可以在一个进程中同时运行Node.js和.NET代码.在本文中,我将会论述这个项目背后的动机,并描述Edge.js提供的基本机制.随后将探讨一些Edge.js应用场景,它在这 ...

  2. Node.js Web 开发框架大全《中间件篇》

    这篇文章与大家分享优秀的 Node.js 中间件模块.Node 是一个服务器端 JavaScript 解释器,它将改变服务器应该如何工作的概念.它的目标是帮助程序员构建高度可伸缩的应用程序,编写能够处 ...

  3. Node.js 入门手册:那些最流行的 Web 开发框架

    这篇文章与大家分享最流行的 Node.js Web 开发框架.Node 是一个服务器端 JavaScript 解释器,它将改变服务器应该如何工作的概念.它的目标是帮助程序员构建高度可伸缩的应用程序,编 ...

  4. Understanding Asynchronous IO With Python 3.4's Asyncio And Node.js

    [转自]http://sahandsaba.com/understanding-asyncio-node-js-python-3-4.html Introduction I spent this su ...

  5. Node.js timer的优化故事

    前几天nodejs发布了新版本4.0,其中涉及到一个更新比较多的模块,那就是下面要介绍的timer模块. timers: Improved timer performance from porting ...

  6. KoaHub.JS用于Node.js的可移植Unix shell命令程序代码

    shelljs Portable Unix shell commands for Node.js ShellJS - Unix shell commands for Node.js     Shell ...

  7. (译+注解)node.js的C++扩展入门

    声明:本文主要翻译自node.js addons官方文档.部分解释为作者自己添加. 编程环境: 1. 操作系统 Mac OS X 10.9.51. node.js v4.4.22. npm v3.9. ...

  8. Practical Node.js (2018版) 第9章: 使用WebSocket建立实时程序,原生的WebSocket使用介绍,Socket.IO的基本使用介绍。

    Real-Time Apps with WebSocket, Socket.IO, and DerbyJS 实时程序的使用变得越来越广泛,如传统的交易,游戏,社交,开发工具DevOps tools, ...

  9. Practical Node.js (2018版) 第7章:Boosting Node.js and Mongoose

    参考:博客 https://www.cnblogs.com/chentianwei/p/10268346.html 参考: mongoose官网(https://mongoosejs.com/docs ...

随机推荐

  1. google nmt 实验踩坑记录

       最近因为要做一个title压缩的任务,所以调研了一些text summary的方法.    text summary 一般分为抽取式和生成式两种.前者一般是从原始的文本中抽取出重要的word o ...

  2. Java 中时间处理 System.currentTimeMillis()

    import org.testng.annotations.Test;import java.text.ParseException;import java.text.SimpleDateFormat ...

  3. PHP自带调试函数

    1.var_dump:打印变量的相关信息 $a = array(1, 2, array("a", "b", "c")); var_dump( ...

  4. DownLoadImage

    Private Declare Function URLDownloadToFile Lib "urlmon" Alias "URLDownloadToFileA&quo ...

  5. CF576E Painting Edges

    首先,有一个很暴力的nk的做法,就是对每种颜色分别开棵lct来维护. 实际上,有复杂度与k无关的做法. 感觉和bzoj4025二分图那个题的区别就在于这个题是边dfs线段树边拆分区间.

  6. 『Numpy学习指南』排序&索引&抽取函数介绍

    排序: numpy.lexsort(): numpy.lexsort()是个排字典序函数,因为很有意思,感觉也蛮有用的,所以单独列出来讲一下: 强调一点,本函数只接受一个参数! import nump ...

  7. 3-1 LVS-NAT集群

    ---- (整理)By 小甘丶 什么是集群: 集群是一组相互独立的.通过高速网络互联的计算机,它们构成了一个组,并以单一系统的模式加以管理.(Cluster就是一组计算机,它们作为一个整体向用户提供一 ...

  8. priority_queue与bfs不得不说的古寺

    前几天写到bfs,看到之前写的,突然感觉不对,后来发现自己把点权值默认当成了边权值,导致一直走不出来: 点权值嘛,就是经过这个点时,要付出这么多的代价,边权值则是经过边时付出,二者有区别滴: 边权值求 ...

  9. CodeForces 558B

    Description Amr has got a large array of size n. Amr doesn't like large arrays so he intends to make ...

  10. POJ-3414 Pots (BFS)

    Description You are given two pots, having the volume of A and B liters respectively. The following ...