JAVA语言程序设计课后习题----第五单元解析（仅供参考）

1　　　　本题是水题，题目要求你求最大值、最小值，建议你用Arrays.sort函数进行排序，最大值、最小值就可以确定了

 import java.util.Arrays;
 import java.util.Scanner;

 public class paone {
     public static void main(String[] args) {
        int [] aa = new int[5];
        double sum = 0.0;
        Scanner input =new Scanner(System.in);
         for (int i = 0; i <5 ; i++) {
             aa[i] = input.nextInt();
             sum += aa[i];
         }
         Arrays.sort(aa);
         for (int a : aa)
             System.out.print(a+" ");
         System.out.println();
         System.out.println("最小值:"+aa[0]);
         System.out.println("最大值:"+aa[4]);
         System.out.println("平均值:"+sum/5);
     }
 }

2　　本题建议用swtich函数进行统计1-6的整数，每次产生一个随机数，就进行+1，switch中进行统计数目的大小建议用素组来定义。

 import java.util.Random;

 public class potwo {
     public static void main(String[] args) {
         Random r = new Random();
         int [] a =new int[1000];
         int [] b =new int[6];
         int count = 0;
         for (int i = 0; i < 1000; i++) {
             a[i] = r.nextInt(6)+1;
             switch (a[i]){
                 case 1:
                     b[0]++;
                     break;
                 case 2:
                     b[1]++;
                     break;
                 case 3:
                     b[2]++;
                     break;
                 case 4:
                     b[3]++;
                     break;
                 case 5:
                     b[4]++;
                     break;
                 case 6:
                     b[5]++;
                     break;
             }
         }
         for (int A : a){
             count++;
             System.out.print(A+" ");
             if (count % 10 == 0)
                 System.out.println();
         }
         System.out.println("1的个数为"+b[0]);
         System.out.println("2的个数为"+b[1]);
         System.out.println("3的个数为"+b[2]);
         System.out.println("4的个数为"+b[3]);
         System.out.println("5的个数为"+b[4]);
         System.out.println("6的个数为"+b[5]);
     }
 }

3　　本题水题，还是用sort函数进行排序，直接找到最小元素

 import java.util.Arrays;
 import java.util.Scanner;

 public class pothree {
     public static void main(String[] args) {
         double [] arrays = new double[5];
         Scanner input = new Scanner(System.in);
         for (int i = 0; i <5 ; i++) {
             arrays[i]=input.nextDouble();
         }
         System.out.println("这个元素最小值是:"+min(arrays));
     }
     public static double min(double[] array){
         Arrays.sort(array);
         return array[0];
     }
 }

4　　本题水题根据题目意思写即可，注意交换位置如何交换 a ^= b;　b ^= a;　a ^= b;

 import java.util.Scanner;

 public class pofour {
     public static void main(String[] args) {
         int [] aa = new int[10];
         Scanner input =new Scanner(System.in);
         for (int i = 0; i < 10 ; i++) {
             aa[i] = input.nextInt();
         }
         System.out.println("排序前的元素:");
         for (int A : aa)
             System.out.print(A+" ");
         System.out.println();
         for (int i = 0; i < 4 ; i++) {
             aa[i] ^= aa[9-i];
             aa[9-i] ^= aa[i];
             aa[i] ^= aa[9-i];
         }
         System.out.println("交换后的元素:");
         for (int A : aa)
             System.out.print(A+" ");
     }
 }

5　　理解题目意思即可

 import java.util.Scanner;

 public class pofive {
     public static void main(String[] args) {
         int [] aa = new int[8];
         Scanner input =new Scanner(System.in);
         for (int i = 0; i < 8; i++) {
             aa[i]=input.nextInt();
         }
         System.out.println("选择排序前:");
         for (int A : aa)
             System.out.print(A+" ");
         System.out.println();
         for (int i = 0; i < 7; i++) {
             int k = i;
             for (int j = k+1; j < 8 ; j++) {
                 if (aa[j] < aa[k])
                     k = j;
             }

             if (i!=k) {
                 aa[i] ^= aa[k];
                 aa[k] ^= aa[i];
                 aa[i] ^= aa[k];
             }
            /* for (int A : aa)
                 System.out.print(A+" ");
             System.out.println();*/
         }
         System.out.println("选择排序后:");
         for (int A : aa)
             System.out.print(A+" ");
     }
 }

6　　根据题目意思，把Fibonacci数列意思理解即可

 public class posix {
    public int [] aa = new int[20];
     public static void main(String[] args) {
         posix posix =new posix();
         int i,count=0;
         for (i = 0; i < 20; i++) {
             System.out.print(posix.fic(i)+" ");
             count++;
             if (count % 10 ==0)
                 System.out.println();
         }

     }
     public int fic(int n){
         if (n==0)
             return aa[n]=1;
         if (n==1)
             return aa[n]=1;
         else
             return fic(n-2)+fic(n-1);
     }
 }

7　　本题有点坑，你写的可能只是一种结果，因为你定义a、b两个数组，两个数组的长度有三种可能 a>b;a=b;a<b;三种可能，注意即可

 import java.util.Scanner;

 public class Poseven {
     public static void main(String[] args) {
         //int [] aa ={1,2,3,4,15};
         //int [] bb ={4,5,6,7};

         int a,b;
         Scanner input = new Scanner(System.in);

         System.out.print("请输入第一个数组的元素个数：");
         a=input.nextInt();
         System.out.print("请输入第二个数组的元素个数：");
         b=input.nextInt();
         int []aa=new int[a];
         int []bb=new int[b];
         System.out.println("请输入第一个数组的元素：");
         for (int i = 0; i < a; i++) {
           aa[i] = input.nextInt();
         }
         System.out.println("请输入第二个数组的元素：");
         for (int i = 0; i < b; i++) {
             bb[i] = input.nextInt();
         }
         System.out.println("加之前两个数组分别为:");
         for (int AA : aa)
             System.out.print(AA+" ");
         System.out.println();
         for (int BB : bb)
             System.out.print(BB+" ");
         System.out.println();
         System.out.println("相加后数组为:");
         Poseven poseven =new Poseven();
        int []abs= poseven.sumAraay(aa,bb);
        for (int ABS : abs)
            System.out.print(ABS+" ");

     }
     public static int[] sumAraay(int [] a,int [] b){
         int max,min;
         max = a.length >= b.length ? a.length:b.length;
         min = a.length <  b.length ? a.length:b.length;
         int leng1=a.length,leng2=b.length;
         int sum []=new int[max];
         for (int i = 0; i < min; i++) {
             sum[i] = a[i]+b[i];
             leng1--;
             leng2--;
         }
         for (int j = min; j < max; j++) {
             if (leng1==0)
                 sum[j] = b[j];
             else
                 sum[j] = a[j];
         }
        /* for (int SUM : sum)
             System.out.print(SUM+" ");
         System.out.println();*/
         return sum;

     }

 }

8　　定义一个可以放下这两个素组大小的素组，记录一个数组存完后，这个素组的下标即可

 import java.util.Arrays;

 public class Poeight {
     public static void main(String[] args) {
         Poeight poeight =new Poeight();
         int [] aa = {1,10,3,4};
         int [] bb = {4,5,6,7,8};
         System.out.println("和成前的数组分别为:");
         for (int AA : aa)
             System.out.print(AA+" ");
         System.out.println();
         for (int BB : bb)
             System.out.print(BB+" ");
         System.out.println();
         System.out.println("合成后的数组为:");
         int [] merge=poeight.arrayMerge(aa,bb);
         for (int MERGE : merge)
             System.out.print(MERGE+" ");
         System.out.println();

     }
     public static int [] arrayMerge(int [] a,int [] b){
         int sum=a.length+b.length;
         int [] suml = new int[sum];
         int min = a.length < b.length ? a.length:b.length;
         int max = a.length >= b.length ? a.length:b.length;
         for (int i = 0; i < min ; i++) {
             if (a.length < b.length)
             suml[i] = a[i];
             else
                 suml[i] = b[i];
         }
         for (int i = min; i < sum; i++) {
             if (a.length>b.length)
             suml[i] = a[i-b.length];
             else
                 suml[i] = b[i-a.length];
         }
         Arrays.sort(suml);
         return suml;
     }
 }

9　　根据题目意思，直接求解即可

 import java.util.Scanner;

 public class Ponine {
     public static void main(String[] args) {
         double [] aa = new double[3];
         double [] bb = new double[3];
         Scanner input = new Scanner(System.in);
         System.out.print("请依次输入a,b,c的值:");
         aa[0] = input.nextDouble();
         aa[1] = input.nextDouble();
         aa[2] = input.nextDouble();
         double x1 =( -aa[1]+Math.sqrt(Math.pow(aa[1],2)-4*aa[0]*aa[2]))/2*aa[0];
         double x2 =( -aa[1]-Math.sqrt(Math.pow(aa[1],2)-4*aa[0]*aa[2]))/2*aa[0];
         bb[0]=x1;
         bb[1]=x2;

         System.out.println("根的个数为"+ solveQuadratic(aa,bb));

     }
     public static int solveQuadratic(double [] eqn, double [] roots){
         int solve=-1;
         if(Math.pow(eqn[1],2)-4*eqn[0]*eqn[2]==0)
             solve=1;
         if(Math.pow(eqn[1],2)-4*eqn[0]*eqn[2] > 0)
             solve=2;
         return solve;
     }
 }

10　　本题有个小技巧，我用的是“置0法”，把不符合要求的全部置0，2是素数，所以我们从下标为1的开始筛选，这个方法有点坑，因为你这样筛选可能会把本身给筛选掉，就需要错位筛选，比如筛选2的倍数，我们从3开始筛选，依次类推，因为求的是2~100的素数，所以因子只可能到50，定义循环的次数即可

 public class Poten {
     public static void main(String[] args) {
         int a[] = new int[100];
         //int b[] =new int[99];
         int i,k,j,cout=0;
         for (i = 0; i < 99 ; i++) {
             a[i] = i+2;
         }
         for (int A : a)
             System.out.print(A+" ");
         System.out.println();
        /* System.out.println();
         k=0;
         System.out.println(a.length);
         for (i=0;i<a.length;i++) {
             if (a[i] % 2 != 0) {
                 b[k] = a[i];
                 k++;
             }
         }
         for (i=0;i<b.length;i++)
             System.out.print(b[i]+" ");*/
 //          把每次结果都输出来显示。
         for (j = 2; j <=50 ; j++) {
             for (k = j-1 ; k<a.length ; k++) {
                 if (a[k] % j == 0)
                     a[k]=0;
             }
             for (int A : a)
                 System.out.print(A+" ");
             System.out.println();
         }

         for (i=0;i<a.length;i++){
             if (a[i]!=0)
                 System.out.print(a[i]+" ");
         }
        // a[i] != '\0'
         //System.out.println(cout);
     }
 }

11　　根据题目意思理解即可

 import java.util.Arrays;

 public class Poeleven {
     public static void main(String[] args) {
         int [] a={1,2,3,4};
         int [] b={1,2,3,4};

        System.out.println( equals(a,b));
        // System.out.println(a.equals(b));
     }
     public static boolean equals (int [] list1,int []  list2){
 //        定义count的原因是把满足题目条件的结果进行加一
         int count=0;
         if(Arrays.equals(list1,list2)){
             for (int i = 0; i <list1.length ; i++) {
                if(list1[i]==list2[i])
                    count++;
             }
         }
 //        满足条件的结果与数组长度相等则满足
         if (count==list1.length)
             return true;
         else
             return false;
     }

 }

12　　本题用的 较简单的方法与第十题基本想法一致，每次循环一遍就把那个数“置0”

 public class Potwleven {

     public static void main(String[] args) {
         int j=0,i=0;
         int a[]={1,2,3,4,5,6,7,8,9,10,11,12};
        //int t= s(a);
 //        while循环的条件就是数组里面不为18的数大于1
         while (s(a)>1){
 //            因为是12个数，则用%12来进行循环
             if (a[i%12]!=18)
                 j++;
             if (j % 5 == 0)
                 a[i%12]=18;
             i++;
         }
        // System.out.println(j);
         for (int k = 0; k <a.length ; k++) {
             if (a[k]!=18)
                 System.out.println(a[k]);
         }
     }
     public static int s(int a[]){
         int count=0;
         for (int i = 0; i < a.length; i++) {
             if (a[i]!=18)
                 count++;
         }
         return count;
     }
 }

13　　本题的选牌次数纯属运气，因为有随机，根据题目意思定义一个二维数组来存放牌

 import java.util.Random;

 public class threeteen {
     public static void main(String[] args) {
         int [][]desk = new int[4][13];
         int count=1;
 //        用来存放选牌的次数
         int cu=0;
         boolean flag =true;
         for (int i = 0; i <4 ; i++) {
             for (int j = 0; j < 13; j++) {
                 desk[i][j]=count;
                 count++;
             }
             count=1;
         }

         Random random =new Random();
         while (flag) {
             cu++;
             int r1 = random.nextInt(4);
             int r2 = random.nextInt(13);
             int a = desk[r1][r2];
             int b = desk[r1][r2];
             int c = desk[r1][r2];
             int d = desk[r1][r2];
             if (a+b+c+d==24)
                 flag=false;

         }
         System.out.println(cu);
     }

    /* public static void main(String[] args) {
         int[] sum = new int[52];
         int[] a = new int[4];
         boolean flag = true;
         for (int i = 0; i < 52; i++) {
             sum[i] = i % 13 + 1;
         }
         for (int i = 0; i < sum.length; i++) {
             System.out.print(sum[i] + " ");
         }
         System.out.println();

         Random random = new Random();
         int r = random.nextInt(52);
         int n = sum.length;
         int cu = 0;
         while (flag) {
             for (int i = 0; i < 4; i++) {
                 a[i] = sum[r];
                 for (int j = r; j < n - 1; j++) {
                     sum[j] = sum[j + 1];
                 }
                 n--;
                 for (int j = 0; j < n; j++) {
                     System.out.print(sum[j] + " ");
                 }
                 System.out.println();
             }

             System.out.println();
             if (a[0]+a[1]+a[2]+a[3]==24)
                 flag=false;
           }*/

 }

14　　本题就是求因子，每次求得的因子就让他进栈，知道最后一次进栈，最后再出栈，即可

 import java.util.Scanner;

 public class Pofourteen {
     private int[]elements;
     private int size=0;
     public static final int DEFAULT_CAPACITY = 10;

     public Pofourteen(){
         this(DEFAULT_CAPACITY);
     }
     public Pofourteen(int capacity){
         elements = new int[capacity];
     }
     //进栈
     public void push(int value){
         if(size >= elements.length){
             int []temp = new int[elements.length*2];
             System.arraycopy(elements,0,temp,0,elements.length);
             elements = temp;
         }
         elements[size++]=value;
     }
     //出栈
     public int pop() {
         return elements[--size];
     }

     public int peek() {
         return elements[size--];
     }
 //  栈是否为空
     public boolean empty() {
         return size == 0;
     }

     public int getSize() {
         return size;
     }

     public static void main(String[] args) {
         Scanner input = new Scanner(System.in);
         int number = input.nextInt();
         Pofourteen stack = new Pofourteen();
         Pofourteen stack1 =new Pofourteen();
         int ans=0;
         int[] a = new int[20];
         while (number != 1) {
             for (int i = 2; i <= number; i++) {
                 if (number % i == 0) {
                    // a[ans] = i;
                     stack.push(i);
                     ans++;
                     number /= i;
                     break;
                 }
             }
         }
         while (!stack.empty()) {
             stack1.push(stack.pop());
          // System.out.print(stack1.pop() + " ");
         }
         //System.out.println();
         while (!stack1.empty()) {
            // stack1.push(stack.pop());
             System.out.print(stack1.pop() + " ");
         }
     }
 }

15　　注意转置是什么意思，就是行列里面的元素互换

 public class Pofifteen {
     public static void main(String[] args) {
         int [][]A= {{1,3,5},{-3,6,0},{13,-5,7},{-2,19,25}};
         int [][]B={{0,-1,2},{7,-1,6},{-6,13,2},{12,-8,-13}};
         System.out.println("A+B矩阵");
         jia(A,B);
         System.out.println("A-B矩阵");
         jian(A,B);
        // System.out.println(A[0].length);
       //  System.out.println(A.length);
         System.out.println("*********************");
         System.out.println("A的转置");
         zhuan(A);
     }
     public static void jia(int [][]a,int [][]b){
         int [][] add= new int[4][3];
         for (int i = 0; i < 4; i++) {
             for (int j = 0; j < 3; j++) {
                 add [i][j] = a[i][j]+b[i][j];
                 System.out.print(add [i][j]+" ");
             }
             System.out.println();
         }
     }
     public static void jian(int [][]a,int [][]b){
         int [][] add= new int[4][3];
         for (int i = 0; i < 4; i++) {
             for (int j = 0; j < 3; j++) {
                 add [i][j] = a[i][j]-b[i][j];
                 System.out.print(add [i][j]+" ");
             }
             System.out.println();
         }
     }
     public static void zhuan(int [][]a){
         int [][]AA=new int[3][4];
         for (int i = 0; i < a[0].length; i++) {
             for (int j = 0; j < a.length; j++) {
                 AA[i][j]=a[j][i];
             }
         }
         for (int i = 0; i < a[0].length; i++) {
             for (int j = 0; j < a.length; j++) {
                 System.out.print(AA[i][j]+" ");
             }
             System.out.println();
         }
     }
 }

16　　把二维素组的下标存入数组即可

 import java.util.Scanner;

 public class Posixteen {
    public static int a1,a2;
     public static void main(String[] args) {
         System.out.print("请输入数组的行数和列数:");
         Scanner input = new Scanner(System.in);

         a1 = input.nextInt();
        a2 =input.nextInt();
        double [][]A=new double[a1][a2];
         for (int i = 0; i <a1 ; i++) {
             for (int j = 0; j <a2 ; j++) {
                 A[i][j] = input.nextDouble();
             }
         }
         int C[]=locateLargest(A);
         System.out.println("最大数的位置是:"+"("+C[0]+","+C[1]+")");
     }
     public static int[] locateLargest(double [][] a){
        int n=0;
        int m=0;
        int c[]=new int[2];
         for (int i = 0; i <a1 ; i++) {
             for (int j = 0; j <a2 ; j++) {
              //   a[i][j] = input.nextDouble();
                 if (a[i][j]>a[n][m]) {
                     c[0] = i;
                     c[1] = j;
                     n=i;
                     m=j;
                 }
             }
         }
         return c;
     }
 }

17　　本题水题，但是有点意思，根据题目的意思写即可

 import java.util.Scanner;

 public class Potwelve {
    /* public static void main(String[] args) {
         Scanner input = new Scanner(System.in);
         int num = input.nextInt();
         int count = 1;
         int n=0,m=num/2;
         int temp1=0,temp2=0;
         int [][]A = new int[num][num];
         for (int i = 0; i <num*num; i++) {

             //    System.out.println(n+" "+m);

              //   if (A[n][m]==0){
                   //  A[n][m]=count;
             if (A[n][m]!=0) {
                 n=++temp1;
                 m=temp2;
             }
                 A[n][m]=count;

                  //   System.out.println(n+" "+m+" "+A[n][m]);
                     count++;
                     temp1 = n;
                     temp2 =m;
                     n--;
                     m++;
                     if (n==-1)
                         n=num-1;
                     if (m==num)
                         m=0;
                // }
         }
         for (int i = 0; i <num; i++) {
             for (int j = 0; j <num; j++) {
                 System.out.print(A[i][j]+" ");
             }
             System.out.println();
         }
     }*/
    public static void main(String[] args) {
        Scanner input =new Scanner(System.in);
        int num = input.nextInt();
        int [][]A=new int[num][num];

        int x=0,y=num/2;
        for (int i = 1; i <=num*num ; i++) {

            if (A[x][y]!=0) {
                //A[x][y] = i;
                x=(x+2)%num;
              //  y=(y+1)%num;
 //               保存原来位置的y
                y=(y+num-1)%num;
            }
           A[x][y]=i;
            x=(x+num-1)%num;
            y=(y+1)%num;
        }
        for (int i = 0; i < num; i++) {
            for (int j = 0; j < num; j++) {
                System.out.print(A[i][j]+" ");
            }
            System.out.println();
        }
    }
 }