Question

我正在编写一个简单的程序来计算2的幂。用户输入他们想要计算2的幂的次数，假设用户输入4，我的程序将返回2,4,8，这是代码

import java.util.Scanner;

public class PowersOf2

{

public static void main(String[] args)

{

int numPowersOf2;        
//How many powers of 2 to compute

int nextPowerOf2 = 1;    
//Current power of  2

int exponent = 0;            

//Exponent for current power of 2 -- this

//also serves as a counter for the loop

Scanner scan = new Scanner(System.in);

        System.out.println("How many powers of 2 would you like printed?");
        numPowersOf2 = scan.nextInt();

        //print a message saying how many powers of 2 will be printed
        //initialize exponent -- the first thing printed is 2 to the what?

    System.out.println("Here are the first " + numPowersOf2 + " power of 2");


      while (exponent<numPowersOf2)
        {
          //print out current power of 2

            nextPowerOf2=nextPowerOf2*2;

            exponent++;

            System.out.println(nextPowerOf2);
            //find next power of 2 -- how do you get this from the last one?

            //increment exponent
             }
        }
}

如果我希望它从0开始首先说2 ^ 0 = 1，那么如果用户输入4，它将返回1,2,4,8而不是2,4,8,16。如何修改它以获得它？

Answer 1

2 ⁿ ==（1 <＆lt; n）; 0＆lt; = n＆lt; 0 32

Answer 2

只需在更改之前打印nextPowerOf2。

while (exponent<numPowersOf2)
{
    System.out.println(nextPowerOf2);

    nextPowerOf2=nextPowerOf2*2;
    exponent++;
}

Answer 3

只能在乘以2之前打印该值。

 //print out current power of 2
 System.out.println(nextPowerOf2);
 nextPowerOf2=nextPowerOf2*2;
 exponent++;

Answer 4

首先打印，然后更新。只需重新安排操作：

while (exponent<numPowersOf2)
{
    //print out current power of 2
    System.out.println(nextPowerOf2);

    //find next power of 2
    nextPowerOf2=nextPowerOf2*2;

    //increment exponent
    exponent++;
}

现在在合理的地方也有正确的缩进和评论！

顺便说一下，使用for循环编写它更有意义：

for (int exponent = 0; exponent < numPowersOf2; ++exponent) {
    //print out current power of 2
    System.out.println(nextPowerOf2);

    //find next power of 2
    nextPowerOf2=nextPowerOf2*2;
}

Answer 5

 nextPowerOf2 = 1;
 exponent     = 0;
 do
 {
     System.out.println(nextPowerOf2);
     nextPowerOf2=nextPowerOf2*2;
     exponent++;
 }
 while( exponent < numPowersOf2 );

应该这样做。

Answer 6

试试这些？它们速度非常快，适合大多数前期，当前和下一个需求。

编辑：注意到你希望它打印出所有，你可以做的是while循环添加打印输出。

public static int getPowerOfTwo(int size)
{
    int n = -1;
    while (size >> ++n > 0);
    return (1 << n - 1 == size) ? size : 1 << n;
}

public static int getNextPowerOfTwo(int size)
{
    int n = -1;
    while (size >> ++n > 0);
    return 1 << n;
}

public static int getPreviousPowerOfTwo(int size)
{
    int n = -1;
    while (size >> ++n > 0);
    return 1 << n - 1;
}

万一你无法弄明白，下面的内容应该有效。

public static void displayNextPowersOfTwo(int size)
{
    int n = -1;
    while (size >> ++n > 0) {
         System.out.print((1 << n) + ",");
    }
    System.out.print('\n');
}

Answer 7

int numpowerof2 = 4, nextpower0f2 = 2, exponent = 0;
while(exponent < numpowerof2) {
  System.out.println(nextpower0f2);
  nextpower0f2 = nextpower0f2 * 2;
  exponent++;
}

Answer 8

而不是使用：

while (exponent < numPowersOf2)

使用此：

while (exponent <= numPowersOf2)

// <= will start with 0, which is 2^0=1

程序计算能力为2

8 个答案: