package sizeyunsuan;
//import java.util.Scanner;
//import java.util.Random;
import java.util.*;
public class jisuan {


public static void count2(int n1[],int n2[],int i,String k)
{
System.out.println(n1[i]+ k +n2[i] +"=" );
}

public static void count3(int n1[],int n2[],int n3[],int i,String k,String t,int brackets)
{
switch(brackets) {
case 0:System.out.println("("+n1[i]+k+n2[i]+")"+t+n3[i]);
case 1:System.out.println(n1[i]+k+"("+n2[i]+t+n3[i]+")");
}

}

public static void twoCount(int i,String k1,int a,int x,int n1[],int n2[])
{
String n="Y";
if(k1.equals(n)) {
switch(a)
{
case 0:
System.out.println(n1[i]+ "+" +n2[i] +"=" );
break;
case 1:
System.out.println(n1[i]+ "-" +n2[i] +"=" );
break;
case 2:
System.out.println(n1[i]+ "*" +n2[i] +"=" );
break;
case 3:
System.out.println(n1[i]+ "/" +n2[i] +"=" );
break;
default:
break;
}
}
else
{
switch(x)
{
case 0:
count2(n1,n2,i,"+");
break;
case 1:
count2(n1,n2,i,"-");
break;
}

}
}

public static void threeCount(int i,String k1,int q,int x,int a,int n1[],int n2[],int n3[],int brackets)
{
String w="Y";
if(k1.equals(w)) {
switch(q)
{
case 0:
count3(n1,n2,n3,i,"+","+",brackets);
break;
case 1:
count3(n1,n2,n3,i,"-","+",brackets);
break;
case 2:
count3(n1,n2,n3,i,"*","+",brackets);
break;
case 3:
count3(n1,n2,n3,i,"/","+",brackets);
break;
case 4:
count3(n1,n2,n3,i,"+","-",brackets);
break;
case 5:
count3(n1,n2,n3,i,"-","-",brackets);
break;
case 6:
count3(n1,n2,n3,i,"*","-",brackets);
break;
case 7:
count3(n1,n2,n3,i,"/","-",brackets);
break;
case 8:
count3(n1,n2,n3,i,"+","*",brackets);
break;
case 9:
count3(n1,n2,n3,i,"-","*",brackets);
break;
case 10:
count3(n1,n2,n3,i,"*","*",brackets);
break;
case 11:
count3(n1,n2,n3,i,"/","*",brackets);
break;
case 12:
count3(n1,n2,n3,i,"+","/",brackets);
break;
case 13:
count3(n1,n2,n3,i,"-","/",brackets);
break;
case 14:
count3(n1,n2,n3,i,"*","/",brackets);
break;
case 15:
count3(n1,n2,n3,i,"/","/",brackets);
break;
default:
break;
}
}
else
{
switch(a)
{
case 0:
count3(n1,n2,n3,i,"+","+",brackets);
break;
case 1:
count3(n1,n2,n3,i,"-","-",brackets);
break;
case 2:
count3(n1,n2,n3,i,"+","-",brackets);
break;
case 3:
count3(n1,n2,n3,i,"-","+",brackets);
break;
}

}
}
public static void main(String[] args)
{
int n,b,e=0,p,number;
String k1,k2;
System.out.println("您需要算题的数量:");
Scanner s=new Scanner(System.in);
n=s.nextInt();


System.out.println("您需要操作数的个数:提示2或3");
number=s.nextInt();


System.out.println("您是否需要乘法和除法:请输入Y或N");
Scanner c=new Scanner(System.in);
k1=c.nextLine();



System.out.println("请输入您可接受的最大数值");
p=s.nextInt();



System.out.println("您是否需要括号介入计算:请输入Y或N");
k2=c.nextLine();
String w="N";

int[] n1=new int[n];
int[] n2=new int[n];
int[] n3=new int[n];
int a;
int q;
int x;
int brackets;
Random random = new Random();
for(int i = 0;i < n;i++)
{
q = random.nextInt(16);
a = random.nextInt(4);
x = random.nextInt(2);
n1[i] = random.nextInt(p);
n2[i] = random.nextInt(p);
n3[i] = random.nextInt(p);
brackets=random.nextInt(2);
/*for(b=0;b<i;b++)
{
if(n1[b]==n1[i]&&n2[b]==n2[i]&&a[b]==a[i]&&x[b]==x[i]&&n3[b]==n3[i])
{
e=1;
}
}*/
if(e==1) continue;
if(number==2&&k2.equals(w))
{
twoCount(i,k1,a,x,n1,n2);
}

else
{
threeCount(i,k1,q,x,a, n1, n2, n3, brackets);
}
}
}
}

可能这道题对于一般人不会特别的难为,但是对于还处在开始的我,可谓举步维艰,一波三折。将近5个小时的时间量,几乎可以达到标准。希望这是个开始......

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