四则运算初级版
package 课堂测验;
import sun.lwawt.macosx.CSystemTray;
import java.util.Arrays;
import java.util.Random;
import java.util.Scanner;
public class operator1 {
static int n;
static int m;
static int chengchu;
static int kuohao;
static int fanwei;
static int count;
public static boolean ifSame(int shu[])
{
Arrays.sort(shu);
for (int i = 0; i < shu.length-1; i++)
{
if(shu[i]==shu[i+1])
return false;
}
return true;
}
public static void canshu()
{
Random r = new Random();
Scanner sc = new Scanner(System.in);
System.out.println("请输入打印数量");
n = sc.nextInt();
System.out.println("请输入打印方式（列数）");
m = sc.nextInt();
System.out.println("请选择是否有乘除法（输入1有，输入2没有）");
chengchu = sc.nextInt();
System.out.println("请选择是否有括号（输入1有，输入2没有）");
kuohao=sc.nextInt();
System.out.println("请输入操作数的范围");
fanwei = sc.nextInt();
System.out.println("请输入操作数的个数");
count = sc.nextInt();
}
public static void dati()
{
}
public static void main(String[] args) {
canshu();
Random r=new Random();
for (int i = 0; i < n; i++)
{
int flag=0;
int czshu[] = new int[count];
Random r1=new Random();
int zkh=r1.nextInt(count-1);//左括号的位置
int ykh=r1.nextInt(count)%(count-zkh)+zkh+1;//右括号的位置
for (int j = 0; j <count-1; j++)
{
czshu[j]=r.nextInt(fanwei)+1;
String fuhao = "";
if(chengchu==1)
{
int f = r.nextInt(4);
if (f == 0) {
fuhao = "+";
} else if (f == 1) {
fuhao = "-";
} else if (f == 2) {
fuhao = "*";
} else if (f == 3) {
fuhao = "/";
}
}
else if(chengchu==2)
{
int f = r.nextInt(2);
if (f == 0) {
fuhao = "+";
} else if (f == 1) {
fuhao = "-";
}
}
if(kuohao==1)
{
if (j == zkh)
{
System.out.print("(");
}
System.out.print(czshu[j]);
if (j == ykh)
{
System.out.print(")");
flag = 1;
}
System.out.print(fuhao);
}
if(kuohao==2)
{
System.out.print(czshu[j]+fuhao);
}
}
int shu[]=new int [n];
if(kuohao==1)
{
if (flag == 0)
{
int s=r.nextInt(fanwei) + 1;
System.out.print(s+ ")=");
shu[i]=s;
}
else
{
int s=r.nextInt(fanwei) + 1;
System.out.print(s+ "=");
shu[i]=s;
}
}
if(kuohao==2)
{
int s=r.nextInt(fanwei) + 1;
System.out.print(s + "=");
shu[i]=s;
}
Scanner sc=new Scanner(System.in);
if ((i + 1) % m == 0)
{
System.out.print("\n");
}
else
{
System.out.print("\t\t");
}
}
}
}
大体上都完成了，除了括号其他都不太难，还有就是不会判断题目是否重复
