随机生成N道四组运算题

Posted 2022-11-21 022414ls

要求：

可随意生成指定数量的题目；可控制每一行有多少个题目；不能有重复；（利用数组查重）

只有加减乘除四则运算，其中乘法要求结果不超过两位数，减法得数不能是负数（换位置），除法结果是整数；

package lianxi1;
import java.util.*;
public class Operation1
 static int i = 0;
 static int j = 0;
 static boolean trag = true;
 static public int getNumbera()
  int a = (int)(1+Math.random()*100);
  return a;
 
 static public int getNumberb()
  int b = (int)(1+Math.random()*100);
  return b;
 
 static public int getNumberc()
  int c = (int)(1+Math.random()*4);
  return c;
 
 

 public static void main(String[] args)
  // TODO Auto-generated method stub
  
  Scanner sc = new Scanner(System.in);
  System.out.println("请输入需要的题目数量：");
  int n = sc.nextInt();
  System.out.println("请输入一行的题目数量：");
  int line = sc.nextInt();
  int x[] = new int[100000];
  int y[] = new int[100000];
  int z[] = new int[100000];
  int temp;
  for(int i=0;i<n;i++)
   for(int j=0;j<line;j++,i++)
    char f = ‘0‘;
    x[i] = getNumbera();
    y[i] = getNumberb();
    z[i] = getNumberc();
    
    int k = 0;
    while(x[k]==x[i]&&y[k]==y[i]&&z[k]==z[i]&&k<=i)
     x[i] = getNumbera();
     y[i] = getNumberb();
     z[i] = getNumberc();
     k++;
    
    
    switch(z[i])
    case 1:
     f = ‘+‘;
     break;
    case 2:
     f = ‘-‘;
     if(x[i]<y[i])
      temp = x[i];
      x[i] = y[i];
      y[i] = temp;
     
     break;
    case 3:
     f = ‘*‘;
     if(x[i]*y[i]>100)
      x[i]=x[i]/10;
      if(x[i]*y[i]>100)
       y[i] = y[i]/10;
      
     
     break;
    case 4:
     f = ‘/‘;
     
     if(x[i]%y[i]!=0&&x[i]>y[i])
      x[i] = x[i] - (x[i]%y[i]);
     
     break;
    
    System.out.print(i+". "+x[i]+f+y[i]+"="+" ");
   
   System.out.println();