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  • 每日总结

    动手动脑 1:通过JDK中的System.out.println()方法来理解函数重载

    借此可以看到,println对应了每一种基本类型都进行了一次重载。

    重载至少需要其参数类型或个数有所变化才能进行重载及调用。

    今天竟然只有一个动手动脑欸,那接下来就是二柱子的故事了,题目要求如下:

    阶段1: 1 定义方法验证题目是否重复;并将出题参数用方法定义。 2 定义随机数生成器方法,根据出题参数(出题个数、操作数的个数、确定操作数的取值范围)生成全部的随机操作数。

    阶段2: 1 定义方法实现在线答题;全部完成后,显示答题结果,输出正确率和错题。 2 定义方法实现错题集、错题重练并记录错题的次数功能。

    阶段3: 增加倒计时功能,完成定时测试功能。

    今日源码:

    复制代码
    public class Totenfour {
    
        @SuppressWarnings("resource")
        public static void main(String[] args) {
            
            Random rand = new Random();
            int numT,numU=0,max=10,min=99;
            int choice=0;
            int[] cp=new int[100];
            int[] cd=new int[100];
            int[] keka=new int[100];
            boolean[] pp=new boolean[100];
            int answer=0;
            int pop=0,poe=0;
            int[] happy=new int[100];
            int[] opea=new int[100];
            Scanner in=new Scanner(System.in);
            System.out.print("请输入题目数");
            numT=in.nextInt();
            System.out.print("请输入最大值");
            max=in.nextInt();
            System.out.print("请输入最小值");
            min=in.nextInt();
            randomSet(cp,cd,numT,numU,max,min);
            for(int i=0;i<numT;i++)
            {
                System.out.print("第"+(i+1)+"题  ");
                choice=rand.nextInt(4);
                switch(choice)
                {
                    case 0:
                        opea[i]=0;
                        keka[i]=cp[i]+cd[i];
                        System.out.print(cp[i]+"+"+cd[i]+"=");
                        answer=in.nextInt();
                        if(answer==keka[i]){pp[i]=true;}
                        else {pp[i]=false;happy[poe]=i;poe++;}
                        break;
                    case 1:
                        opea[i]=1;
                        keka[i]=cp[i]-cd[i];
                        System.out.print(cp[i]+"-"+cd[i]+"=");
                        answer=in.nextInt();
                        if(answer==keka[i]){pp[i]=true;}
                        else {pp[i]=false;happy[poe]=i;poe++;}
                        break;
                    case 2:
                        opea[i]=2;
                        keka[i]=cp[i]*cd[i];
                        System.out.print(cp[i]+"x"+cd[i]+"=");
                        answer=in.nextInt();
                        if(answer==keka[i]){pp[i]=true;}
                        else {pp[i]=false;happy[poe]=i;poe++;}
                        break;
                    case 3:
                        opea[i]=3;
                        keka[i]=cp[i][i];
                        System.out.print(cp[i]+"/"+cd[i]+"=");
                        answer=in.nextInt();
                        if(answer==keka[i]){pp[i]=true;}
                        else {pp[i]=false;happy[poe]=i;poe++;}
                        break;
                }
            }
            pop=curry(pp,numT);
            for(;pop!=1;)
            {
                int k=poe;
                System.out.println("这里还有错题哦,一共"+poe+"道题目");
                for(int i=0;i<k;i++)
                {
                    switch(opea[happy[i]])
                    {
                       case 0:
                           System.out.print("第"+(happy[i]+1)+"题 ");
                           keka[happy[i]]=cp[happy[i]]+cd[happy[i]];
                            System.out.print(cp[happy[i]]+"+"+cd[happy[i]]+"=");
                            answer=in.nextInt();
                            if(answer==keka[happy[i]]){pp[happy[i]]=true;poe--;}
                           break;
                       case 1:
                           System.out.print("第"+(happy[i]+1)+"题 ");
                           keka[happy[i]]=cp[happy[i]]-cd[happy[i]];
                            System.out.print(cp[happy[i]]+"-"+cd[happy[i]]+"=");
                            answer=in.nextInt();
                            if(answer==keka[happy[i]]){pp[happy[i]]=true;poe--;}
                           break;
                       case 2:
                           System.out.print("第"+(happy[i]+1)+"题 ");
                           keka[happy[i]]=cp[happy[i]]*cd[happy[i]];
                            System.out.print(cp[happy[i]]+"x"+cd[happy[i]]+"=");
                            answer=in.nextInt();
                            if(answer==keka[happy[i]]){pp[happy[i]]=true;poe--;}
                           break;
                       case 3:
                           System.out.print("第"+(happy[i]+1)+"题 ");
                           keka[happy[i]]=cp[happy[i]][happy[i]];
                            System.out.print(cp[happy[i]]+"/"+cd[happy[i]]+"=");
                            answer=in.nextInt();
                            if(answer==keka[happy[i]]){pp[happy[i]]=true;poe--;}
                           break;
                    }
                }
                pop=curry(pp,numT);
            }
            System.out.println("恭喜,全部正确啦");
            in.close();
        }
        public static int curry(boolean[] i,int numT) {
            for(int c=0;c<numT;c++)
            {
                if(i[c]==false)
                {
                    return 0;
                }
            }
            return 1;
        }
        public static int check(int[] kp,int [] kk,int i,int s) {
            for(int k=0;k<=50;k++)
            {
                if((kk[k]==i)&&(kp[k]==s))
                {return 0;}
            }
            return 1;
        }
        public static void randomSet(int a[],int b[],int numT,int numU ,int max,int min) {
            Random rand = new Random();
            int r;int p;
            for(int x=0;x<numT;x++)
            {
                r=rand.nextInt(max-min+1) + min;
                p=rand.nextInt(max-min+1) + min;
                int op=check(a,b,r,p);
                for(;op!=1;)
                {
                      r=rand.nextInt(max-min+1) + min;
                   p=rand.nextInt(max-min+1) + min;
                   op=check(a,b,r,p);
                }
                a[x]=r;b[x]=p;
            }
        }
    }
    复制代码

    运行结果如下:

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  • 原文地址:https://www.cnblogs.com/ldy2396/p/14218688.html
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