二柱子四则预算生成进阶
作者：互联网
现在终于解决了随机生成括号的问题，操作数可以支持四个了
1 package practice;
2
3 import java.util.Random;
4 import java.util.Scanner;
5
6 public class add {
7 public static void main(String[] args){
8 char[] fh = new char[] {'+','-','×','÷'};
9 int[][] cy = new int[1000][3];
10 Random r= new Random(1);
11 Scanner sc = new Scanner(System.in);
12 System.out.println("请输入需要产生的题目数量.");
13 int n1 = sc.nextInt();
14 System.out.println("请输入操作个数（1.两个 2.三个 3.四个）");
15 int n2 = sc.nextInt();
16 System.out.println("请输入操作数的取值范围（先下限后上限）：");
17 int n3 = sc.nextInt();
18 int n4 = sc.nextInt();
19 System.out.println("是否添加括号？（1.是 2.否）");
20 int n5 = sc.nextInt();
21 if(n2==1) {
22 if(n5==1)System.out.println("两个操作数加括号就没必要了吧");
23 for(int i=0;i<n1;i++) { //两个操作数
24 int ran1 = r.nextInt(n4-n3+1)+n3;
25 int ran2 = r.nextInt(n4-n3+1)+n3;
26 int ch = r.nextInt(4);
27 int temp=0;
28 for(int j=0;j<i;j++) {
29 if(cy[j][0]==ran1&&cy[j][1]==ch&&cy[j][2]==ran2) {
30 i--;
31 temp=1;
32 }
33 }
34 if(temp==1)break;
35 cy[i][0] = ran1;
36 cy[i][1] = ch;
37 cy[i][2] = ran2;
38 System.out.println(ran1+" "+fh[ch]+" "+ran2);
39 }
40 }
41 else if(n2==2) { //三个操作数
42 if(n5==2) {
43 for(int i=0;i<n1;i++) { //三个操作数 无括号
44 int ran1 = r.nextInt(n4-n3+1)+n3;
45 int ran2 = r.nextInt(n4-n3+1)+n3;
46 int ran3 = r.nextInt(n4-n3+1)+n3;
47 int ch1 = r.nextInt(4);
48 int ch2 = r.nextInt(4);
49 System.out.println(ran1+" "+fh[ch1]+" "+ran2+" "+fh[ch2]+" "+ran3);
50 }
51 }
52 if(n5==1) {
53 for(int i=0;i<n1;i++) { //三个操作数 有括号
54 int temp = 0;
55 int ran1 = r.nextInt(n4-n3+1)+n3;
56 int ran2 = r.nextInt(n4-n3+1)+n3;
57 int ran3 = r.nextInt(n4-n3+1)+n3;
58 int ch1 = r.nextInt(4);
59 int ch2 = r.nextInt(4);
60 char[] kh = new char[6];
61 int k1 = r.nextInt(3)+1;
62 int k2 = r.nextInt(3)+1;
63 if(k2-k1>=1) {
64 kh[2*k1-2]='(';
65 kh[2*k2-1]=')';
66 }else {
67 --i;
68 temp = 1;
69 }
70 if(temp==1)continue;
71 System.out.println(kh[0]+" "+ran1+" "+kh[1]+" "+fh[ch1]+" "+kh[2]+" "+ran2+" "+kh[3]+" "+fh[ch2]+" "+kh[4]+" "+ran3+" "+kh[5]);
72 }
73 }
74 }
75 if(n2==3) {
76 if(n5==2) {
77 for(int i=0;i<n1;i++) { //四个操作数 无括号
78 int ran1 = r.nextInt(n4-n3+1)+n3;
79 int ran2 = r.nextInt(n4-n3+1)+n3;
80 int ran3 = r.nextInt(n4-n3+1)+n3;
81 int ran4 = r.nextInt(n4-n3+1)+n3;
82 int ch1 = r.nextInt(4);
83 int ch2 = r.nextInt(4);
84 int ch3 = r.nextInt(4);
85 System.out.println(ran1+" "+fh[ch1]+" "+ran2+" "+fh[ch2]+" "+ran3+" "+fh[ch3]+" "+ran4);
86 }
87 }
88 if(n5==1) {
89 for(int i=0;i<n1;i++) { //四个操作数 有括号
90 int temp = 0;
91 int ran1 = r.nextInt(n4-n3+1)+n3;
92 int ran2 = r.nextInt(n4-n3+1)+n3;
93 int ran3 = r.nextInt(n4-n3+1)+n3;
94 int ran4 = r.nextInt(n4-n3+1)+n3;
95 int ch1 = r.nextInt(4);
96 int ch2 = r.nextInt(4);
97 int ch3 = r.nextInt(4);
98 char[] kh = new char[8];
99 int k1 = r.nextInt(4)+1;
100 int k2 = r.nextInt(4)+1;
101 if(k2-k1>=1) {
102 kh[2*k1-2]='(';
103 kh[2*k2-1]=')';
104 }else {
105 --i;
106 temp = 1;
107 }
108 if(temp==1)continue;
109 System.out.println(kh[0]+" "+ran1+" "+kh[1]+" "+fh[ch1]+" "+kh[2]+" "+ran2+" "+kh[3]+" "+fh[ch2]+" "+kh[4]+" "+ran3+" "+kh[5]+" "+fh[ch3]+" "+kh[6]+" "+ran4+" "+kh[7]);
110 }
111 }
112 }
113 }
114 }
标签：柱子,进阶,int,kh,System,fh,println,预算,out 来源： https://www.cnblogs.com/SHINIAN200/p/15321821.html