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第三次作業 軟件工程

sys cas 第三次作業 alt args 運算符 spa 個數 還需要

要求0:

作業地址:

https://edu.cnblogs.com/campus/nenu/2016CS/homework/2266

要求1:

git倉庫地址:https://git.coding.net/YUAN-FANG/f4.git

要求2:

結對同學博客地址:https://www.cnblogs.com/xinghl907/

結對同學姓名學號:邢海玲 2016011907

解題思路:

功能1和功能2: 這兩個功能就要求隨機產生N道四則運算題,用戶輸入一個?加答案與標準答案進行比對並判斷對錯,最後統計有多少四則運算題被正確解答。 此次作業從實現四則運算入手,要考慮加減乘除按通常四則運算規定的計算優先級除法用整數除法,即僅保留除法運算結果的整數部分。比如

8/3=2。輸入表達式保證無0作為除數情況發生由於基礎較弱,很多部分尚未完成。

功能的實現:

1.我們運用了三個數組來存放相應的數值以及運算符號:

public static int[] firstNum = new int[100];

public static int[] secondNum = new int[100];

public static String[] mark = new String[]{"+","-","*","/"};

2.下面為簡單代碼實現:

public class SizeYunsuan {

public static int[] firstNum = new int[100];

public static int[] secondNum = new int[100];

public static String[] mark = new String[]{"+","-","*","/"};

public static void main(String[] args){

Random rand = new Random();

Scanner input = new Scanner(System.in);

int number = input.nextInt();

for(int i=0;i<100;i++){

firstNum[i]= rand.nextInt(number);

secondNum[i] = rand.nextInt(number);

}

system.out.println("請輸入產生四則運算的道數:");

number = input.nextInt();

System.out.println("產生四則運算式子:");

for(int j=1;j<=number;j++){

String markYun = mark[rand.nextInt(4)];

int answerInput = 0;

int isFirstNum = firstNum[rand.nextInt(100)];

int isSecondNum = secondNum[rand.nextInt(100)];

switch(markYun){

case "+":

System.out.println(isFirstNum + "+" + isSecondNum + "=" );

answerInput = isFirstNum + isSecondNum;

break;

case "-":

if(isFirstNum > isSecondNum){

System.out.println(isFirstNum + "-" + isSecondNum + "=" );

answerInput = isFirstNum - isSecondNum;

}

else{

System.out.println(isSecondNum + "-" + isFirstNum + "=" );

answerInput = isSecondNum - isFirstNum;

}

break;

case "*":

System.out.println(isFirstNum + "*" + isSecondNum + "=" );

answerInput = isFirstNum * isSecondNum;

break;

case "/":

System.err.println("提示:結果取整數!");

if(isSecondNum != 0){

System.out.println(isFirstNum + "/" + isSecondNum + "=" );

answerInput = isFirstNum / isSecondNum;

}else{

System.out.println(isFirstNum + "/" + secondNum[rand.nextInt(100)]+1 + "=" );

answerInput = isFirstNum / isSecondNum;

}

break;

}

System.out.print("");

int answer = input.nextInt();

while(answer != answerInput){

System.out.print("答案錯誤,請重新輸入:");

answer = input.nextInt();

}

System.out.print("總共\n道題,你答對\n道題!");

}

input.close();

3.代碼運行片段:

技術分享圖片

照片:

技術分享圖片

編程體會:

1.基礎薄弱,還需要努力學習

2.對同一個問題(即使很小的問題)從多個角度思考,能有不同的認識

3.結對編程對程序的質量有明顯的提升

第三次作業 軟件工程