英文:
Flipping a coin and stopping when it lands heads 4 times in a row
问题
任务是反复抛硬币直到连续出现四次正面，然后显示所有导致这种情况的结果。我一直在收到我加入的最后一个错误消息，以防万一出了问题。我不知道我搞错了什么，想知道是否有人能够帮助。
class Main {
public static void main(String[] args) {
int h = 2;
int t = 1;
int count = 0;
int result;
while (count <= 4)
{
result = (int)Math.random() * 2;
if (result == 2)
{
count++;
System.out.print("H ");
}
else if (result == 1)
{
count = 0;
System.out.print("T ");
}
else
System.out.println("error");
}
}
}
英文:
The assignment is to flip a coin until four heads in a row are seen and display all the results leading up to that. I keep getting the last error message I put in just in case it fell through. I have no idea what I messed up and was wondering if someone was able to help.
class Main {
public static void main(String[] args) {
int h = 2;
int t = 1;
int count = 0;
int result;
while (count<=4)
{
result = (int)Math.random()*2;
if (result == 2)
{
count++;
System.out.print("H ");
}
else if (result == 1)
{
count=0;
System.out.print("T ");
}
else
System.out.println("error");
}
}
}
答案1
得分: 2
(int)Math.random() * 2
等同于
((int)Math.random()) * 2
鉴于Math.random()返回至少为零但小于一的数字，您的表达式将始终为零。
加上括号：
(int) (Math.random() * 2)
但是，请同时查看条件语句中“result”的值：您永远不会生成2。
英文:
(int)Math.random() * 2
is the same as
((int)Math.random()) * 2
Given that Math.random() returns a number at least zero but less than one, your expression is always going to be zero.
Put in parentheses:
(int) (Math.random() * 2)
But then, also look at the values of result in your conditionals: you will never generate 2.
答案2
得分: 1
将其中的一部分翻译如下：
你需要加1以获得可能的数值为一或二的情况：
result = (int) (Math.random() * 2 + 1);
英文:
You need to add 1 to have possible values of one or two:
result = (int) (Math.random() * 2 + 1);
答案3
得分: 1
您可以使用 `Random` 类和布尔值：
Random random = new Random();
int count = 0;
while (count < 4) {
if (random.nextBoolean()) {
System.out.print("H");
count++;
} else {
count = 0;
System.out.print("T");
}
}
英文:
You can use the Randomclass and boolean
Random random = new Random();
int count = 0;
while (count < 4) {
if (random.nextBoolean()) {
System.out.print("H");
count++;
} else {
count = 0;
System.out.print("T");
}
}
答案4
得分: 0
由于Math.random()的结果介于0和1之间，将其强制转换为int类型将删除小数点后的数字，你将始终得到零作为答案。
以下代码将帮助您生成在min和max之间的随机数。
// 定义范围
int max = 2;
int min = 1;
int range = max - min + 1;
int rand = (int)(Math.random() * range) + min;
要了解这是如何工作的，您可以将最小可能值设为0，最大可能值设为0.99，并将两者都乘以任何范围，比如说20，答案仍将在1到20.8之间，由于它不是四舍五入而是直接强制转换，所以变为20。因此，这可以为您提供任何范围的随机数。
英文:
As the result of Math.random() is between 0 and 1 type casting it to int will remove the digits after decimal point and you'll always have zero as answer.
Below code will help you to generate a random number between min and max.
// define the range
int max = 2;
int min = 1;
int range = max - min + 1;
int rand = (int)(Math.random() * range) + min;
For an explanation of how this works you can put the min possible value of 0 and max possible of 0.99 and multiply both by any range, let's say 20 the answer will still be in between 1 to 20.8 which gets turned to 20 as it's not rounding off but directly type casting. Hence, this can give you a random number for any range.
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