2016012019——小学四则运算练习软件项目报告

仓库地址：https://git.coding.net/Buza/arithmeticForStu.git

内容一：需求分析

用户用例：

内容二：功能设计

1. 可以输入算式数目
2. 算式结果不可出现小数、负数
3. 生成计算结果正确的三到五个0~100之间正整数的四则运算算式

内容三：设计思路及其结构

·首先这个问题我首先想到的是，可以用栈或者递归来完成，我太菜了，想不出附加题，就用了递归。

·观察算式结构，要想使算式中不出现括号，比较简单的一种方法就是把算式拆分成两部分

·首先构造两个算法，分别生成第一部分的string和在第一部分的基础上逐步拼接第二部分的加减法。同时这两个算法要返回生成计算的结果值。在顾思宇同学的帮助下，我把结果值分别导入了指针为ans[ ]和newAnswer[ ]里面，就可以在不返回数值的同时得到我们需要的结果了。

第一部分：

public String addNum(int a,int b,String aa, int []ans,int d) {
int c;
if(a<b) {
c = b-a;
aa = a+" "+"+"+" "+c;
d = b ;
return aa;
}else if(a>b) {
c = a - b;
aa =  b+" "+"+"+" "+c;
d = a;
return aa;
}else {
aa = a+" "+"+"+" "+0;
d = a;
return aa;
}
}
public String minNum(int a,int b,String aa, int []ans,int d) {

int c;
if(a<b) {
c = b-a;
aa = a+" "+"-"+" "+c;
d = b;
return aa;
}else if(a>b) {
c = b-a;
aa =  b+" "+"-"+" "+c;
d = a ;
return aa;
}else {
aa = a+" "+"-"+" "+b;
d = 0;
return aa;
}
}
public String multNum(int a,int b,String aa, int []ans,int d) {

int c = a*b ;
d = c;
aa = a+" "+"*"+" "+b;
return aa;
}
public String divNum(int a,int b,String aa, int []ans,int d) {

int c = a*b;
if(c<=100) {
aa = c+" "+"%"+" "+a;
d = b;
return aa;
}else {
return divNum(a,b,aa, ans, d);
}

}

其中大家可能都有一个问题，如何避免除法中的分数和减法中的负数。我的思路是，除法生成时，先随机生成一个正整数作为算式的结果，再令这个结果乘一个正整数，得到算式的除数，这样避免了小数的出现；减法中，随机生成的两个数在进行减法时，进行比较，分出减数和被减数就好啦。

第二部分：

public String add(int a,int b,String equ,String bb,int[] newAnswer,int c) {
c = a+b;
bb = equ+" "+"+"+" "+ b;
return bb;
}
public String min(int a,int b,String equ,String bb,int[] newAnswer,int c) {
bb = equ+" "+"-"+" "+ b;
return bb;
}

因为只需要编写加减两部分，就轻松很多。

·最后在返回算式结果时，在新的string中加上=answer（导出的结果值）就好。

·方法构建完了，就要调用它，调用就很难，因为要递归，就感觉很复杂。

调用方法时我们要考虑：

1.需要的重要变量：符号数：numOfSign;  符号选择：（因为用了两个变量，我定义了两个变量来生成随机数）ran、ran0；两个随机正整数：a 、b;和两个数组用于接收结果值：Answer[ ]、newAnswer[ ]。

2.然后就是递归方法了，首先分了三种情况：两个符号、三个符号、四个符号；然后我们得把第二种方法套进去

if(numOfSign == 2) {
int c = (int)(Math.random()*100);
int ran0 = (int)(Math.random()*2);
if(ran == 0) {
addNum(a, b,equations,Answer, 0);
int ans = Answer[0];
if(ran0 == 0) {
add(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}else{
min(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}
}else if (ran == 1) {
minNum(a, b, equations,Answer, 0);
int ans = Answer[0];
if(ran0 == 0) {
add(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}else{
min(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}
}else if (ran == 2) {
multNum(a, b,equations,Answer, 0);
int ans = Answer[0];
if(ran0 == 0) {
add(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}else{
min(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}
}else{
divNum(a, b,equations,Answer, 0);
int ans = Answer[0];
if(ran0 == 0) {
add(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}else{
min(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}
}
int newans = newAnswer[0];
cc = equations+" "+"="+" "+newans;
return cc;
}
//there are two signs above
else if(numOfSign == 3) {
int c = (int)(Math.random()*100);
if(ran == 0) {
addNum(a, b, equations,Answer, 0);
int q;
for(q = 0;q<2;q++) {
int ran0 = (int)(Math.random()*2);
int ans = Answer[0];
if(ran0 == 0) {
add(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}else{
min(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}
}

}else if (ran == 1) {
minNum(a, b, equations,Answer, 0);
int q;
for(q = 0;q<2;q++) {
int ran0 = (int)(Math.random()*2);
int ans = Answer[0];
if(ran0 == 0) {
add(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}else{
min(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}
}
}else if (ran == 2) {
multNum(a, b,equations,Answer, 0);
int q;
for(q = 0;q<2;q++) {
int ran0 = (int)(Math.random()*2);
int ans = Answer[0];
if(ran0 == 0) {
add(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}else{
min(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}
}
}else{
divNum(a, b,equations,Answer, 0);
int q;
for(q = 0;q<2;q++) {
int ran0 = (int)(Math.random()*2);
int ans = Answer[0];
if(ran0 == 0) {
add(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}else{
min(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}
}
}
int newans = newAnswer[0];
cc = equations+" "+"="+" "+newans;
return cc;
}
//there are three signs above
else{
int c = (int)(Math.random()*100);
if(ran == 0) {
addNum(a, b, equations,Answer, 0);
int q;
for(q = 0;q<3;q++) {
int ran0 = (int)(Math.random()*2);
int ans = Answer[0];
if(ran0 == 0) {
add(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}else{
min(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}
}

}else if (ran == 1) {
minNum(a, b, equations,Answer, 0);
int q;
for(q = 0;q<3;q++) {
int ran0 = (int)(Math.random()*2);
int ans = Answer[0];
if(ran0 == 0) {
add(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}else{
min(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}
}
}else if (ran == 2) {
multNum(a, b,equations,Answer, 0);
int q;
for(q = 0;q<3;q++) {
int ran0 = (int)(Math.random()*2);
int ans = Answer[0];
if(ran0 == 0) {
add(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}else{
min(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}
}
}else{
divNum(a, b,equations,Answer, 0);
int q;
for(q = 0;q<3;q++) {
int ran0 = (int)(Math.random()*2);
int ans = Answer[0];
if(ran0 == 0) {
add(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}else{
min(ans, c, equations, equations0,newAnswer,0);
equations = equations0;
}
}
}
int newans = newAnswer[0];
cc = equations+" "+"="+" "+newans;
return cc;
}
//there are four signs above

·最后只需要构造main函数，让用户输入所需题目数，然后用for循环来调用上面的方法。

第四部分：算法讲解

我的算法很烂啊，我感觉没啥说的，但是最近看博客大家提交的都好强啊，我会继续关注大佬们的更新动态的，fighting！

如果一定要说，我感觉递归是个好东西，虽然用不利索。。。。。。

第五部分：测试运行

从周五调到现在了，还是有bug（沮丧.jpg）还没修完，下周接着修，有时候有行，有时候就出问题了，就各种null，应该是某些方法出问题了，目前这样

第六部分：我感觉我除法这里弄的挺好的，这样做就没有分数，但是如果叫我做分数出来，我就不会弄了

public String divNum(int a,int b,String aa, int []ans,int d) {

int c = a*b;
if(c<=100) {
aa = c+" "+"%"+" "+a;
d = b;
return aa;
}else {
return divNum(a,b,aa, ans, d);
}

}

第七部分：总结

模块化很重要的，提前设计程序，把大程序拆成不同功能，面向不同功能的编程可以少走弯路节省很多时间，尤其是我这样的菜菜，做到模块化可以省力，不会很迷茫。模块化最重要的还是：小块修改不会牵扯大块，就很方便了。

第八部分：PSP

上面只是代码部分的PSP。。。。这周末研究coding.net研究到胃疼。最后还是王超学姐帮我找到了问题，总算把git给弄好了，公钥给生成了，可惜我还是不是很懂git的代码，照着教程码是行，但看代码看不懂，不太懂怎么用命令行往移动仓库移东西，还得接着研究呀。