英文:
How do I make a random int?
问题
这里是您的翻译代码部分:
for (i = 1; i <= 100; i++) {
int r = (int) (Math.random() * 100) + 1;
System.out.print(r + ", ");
}
请注意,我只进行了代码的翻译,不包括额外的解释或回答。如果您有任何其他问题或需要进一步的帮助,请随时提问。
英文:
So I am trying to make 100 random integers from 1-100 print out. I can make doubles of this but not integers. Please show me how to make this into integers, here is the code.
for(i = 1; i <= 100; i++) {
double r = Math.random();
double in = r * 100;
System.out.print(in + ", ");
}
Please show me what I messed up or if I need to use an alternate technique for this
答案1
得分: 1
你可以使用 Java.util 包中的 Random 对象替代。
Random random = new Random();
int value = random.nextInt(100);
英文:
You can use Random object instead. from java.util package
Random random = new Random();
int value = random.nextInt(100);
答案2
得分: -1
使用ThreadLocalRandom:
for (int i = 1; i <= 100; i++) {
System.out.print(ThreadLocalRandom.current().nextInt(100) + ", ");
}
英文:
Use ThreadLocalRandom:
for (int i = 1; i <= 100; i++) {
System.out.print(ThreadLocalRandom.current().nextInt(100) + ", ");
}
答案3
得分: -1
导入类 java.util.Random
创建类 Random 的实例,即 Random rand = new Random()
调用 rand 对象的以下方法之一:
nextInt(upperbound) 生成范围在 0 到 upperbound-1 之间的随机数。
nextFloat() 生成介于 0.0 和 1.0 之间的浮点数。
nextDouble() 生成介于 0.0 和 1.0 之间的双精度浮点数
例如,如果您想生成一个介于 1 到 100 之间的随机数,则代码将如下所示。
import java.util.Random;
Random r = new Random();
int number = r.nextInt(100);
因此,变量 number 将是一个介于 1 到 100 之间的随机数。
英文:
Import the class java.util.Random
Make the instance of the class Random, i.e., Random rand = new Random()
Invoke one of the following methods of rand object:
nextInt(upperbound) generates random numbers in the range 0 to upperbound-1.
nextFloat() generates a float between 0.0 and 1.0.
nextDouble() generates a double between 0.0 and 1.0
For example if you want to generate a random number between 1 to 100 the the code will be.
import java.util.Random;
Random r = new Random();
Int number = r.nextInt(100);
So variable number will be a random number between 1 to 100
答案4
得分: -2
以下是翻译好的内容:
错误的是语句 double in = r * 100;
必须是
int b = (int) (Math.random()*100)%100; //范围 [0-99]
如果您严格希望 b 在范围 [1-100],只需将 b 增加 1。
英文:
Whats wrong is the statement double in = r * 100;
It must be
int b = (int) (Math.random()*100)%100; //range [0-99]
Just add 1 to b if you strictly want b in the range [1-100]
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