具有2的倍数的java金字塔不工作
我假设创建一个金字塔,将每个数字乘以2,直到到达中间,然后除以2,如下面的示例所示 但是,在编写代码后,我无法使数字加倍(I*2),然后一旦它到达中心,它就会除以2,直到它变成具有2的倍数的java金字塔不工作,java,for-loop,pyramid,Java,For Loop,Pyramid,我假设创建一个金字塔,将每个数字乘以2,直到到达中间,然后除以2,如下面的示例所示 但是,在编写代码后,我无法使数字加倍(I*2),然后一旦它到达中心,它就会除以2,直到它变成1 我的输出: package question5; public class Pyramid { public static void main(String[] args) { int x = 5; int rowCount = 1; System.out.println("Her
1
我的输出:
package question5;
public class Pyramid {
public static void main(String[] args) {
int x = 5;
int rowCount = 1;
System.out.println("Here Is Your Pyramid");
//Implementing the logic
for (int i = x; i > 0; i--)
{
//Printing i*2 spaces at the beginning of each row
for (int j = 1; j <= i*2; j++)
{
System.out.print(" ");
}
//Printing j value where j value will be from 1 to rowCount
for (int j = 1; j <= rowCount; j++)
{
System.out.print(j+" ");
}
//Printing j value where j value will be from rowCount-1 to 1
for (int j = rowCount-1; j >= 1; j--)
{
System.out.print(j+" ");
}
System.out.println();
//Incrementing the rowCount
rowCount++;
}
}
}
package问题5;
公共阶层金字塔{
公共静态void main(字符串[]args){
int x=5;
int rowCount=1;
System.out.println(“这是你的金字塔”);
//实现逻辑
对于(int i=x;i>0;i--)
{
//在每行开头打印i*2个空格
对于(int j=1;j现在您正在输出j
,一个从1
到5
的数字。您想要输出2j-1,这很简单:1这很有效……您可以使用math.pow方法
public class test {
public static void main(String[] args) {
int x = 5;
int rowCount = 1;
System.out.println("Here Is Your Pyramid");
//Implementing the logic
for (int i = x; i > 0; i--)
{
//Printing i*2 spaces at the beginning of each row
for (int j = 1; j <= i*2; j++)
{
System.out.print(" ");
}
//Printing j value where j value will be from 1 to rowCount
for (int j = 0; j <= rowCount-1; j++)
{
System.out.printf("%2d", (int)Math.pow(2, j));
}
//Printing j value where j value will be from rowCount-1 to 1
for (int j = rowCount-1; j >= 1; j--)
{
System.out.printf("%2d", (int)Math.pow(2, j-1));
}
System.out.println();
//Incrementing the rowCount
rowCount++;
}
}
}
公共类测试{
公共静态void main(字符串[]args){
int x=5;
int rowCount=1;
System.out.println(“这是你的金字塔”);
//实现逻辑
对于(int i=x;i>0;i--)
{
//在每行开头打印i*2个空格
对于(int j=1;j您希望我为loopNo输入吗?只需替换两个System.out.print(j+“”)
好的,谢谢,我现在就这么做,我在代码中放了一张图片,它看起来很奇怪?我在代码中放了一张图片,它看起来很奇怪?这好像你有一个整数序列,你需要把它们映射到一个几何序列上。一旦你找到了公式,你的程序就正确了。
public class test {
public static void main(String[] args) {
int x = 5;
int rowCount = 1;
System.out.println("Here Is Your Pyramid");
//Implementing the logic
for (int i = x; i > 0; i--)
{
//Printing i*2 spaces at the beginning of each row
for (int j = 1; j <= i*2; j++)
{
System.out.print(" ");
}
//Printing j value where j value will be from 1 to rowCount
for (int j = 0; j <= rowCount-1; j++)
{
System.out.printf("%2d", (int)Math.pow(2, j));
}
//Printing j value where j value will be from rowCount-1 to 1
for (int j = rowCount-1; j >= 1; j--)
{
System.out.printf("%2d", (int)Math.pow(2, j-1));
}
System.out.println();
//Incrementing the rowCount
rowCount++;
}
}
}