Question

这是一个非常简单的程序。计算机和用户在0-3之间选择一个数字。如果用户没有猜到与计算机相同的数字，我希望该程序返回。

    String input; // input is a string variable
    int cpucrazy;
    int convertstring;

//第一步//

    input = JOptionPane.showInputDialog("Guess a number between 0-3");
    convertstring = Integer.parseInt(input);

//随机部分//

    Random ran = new Random ();
    cpucrazy = ran.nextInt(3);

//计算！

if (cpucrazy < convertstring) {
     JOptionPane.showInputDialog(null, "Your guess is too high. Guess again?"); }


else if (cpucrazy > convertstring) {
     JOptionPane.showInputDialog(null, "Your guess is too low. Guess again?"); }


else JOptionPane.showMessageDialog(null, "Correct!!");

Answer 1

if (cpucrazy < convertstring) {
 JOptionPane.showInputDialog(null, "Your guess is too high. Guess again?"); }


else if (cpucrazy > convertstring) {
     JOptionPane.showInputDialog(null, "Your guess is too low. Guess again?"); }


else JOptionPane.showMessageDialog(null, "Correct!!");

您需要在上述代码周围放置while循环（或某种类型的循环构造），并在以下情况下退出构造：cpucrazy == convertstring或在cpucrazy != convertstring

停留在循环构造中

循环看起来像下面的伪代码：

b = random();

do
{
     input a;
     if( a != b )
     {
         if(a < b )
         {
              print "Too low";
         }
         else
         {
              print "Too high";
         }
     }
} while( a != b );

print "You got it correct"

再试一次循环循环

1 个答案: