找出前N个五角形数 [英] Find first N pentagonal numbers

问题描述

我必须找到1-100中的第一个N [pentagonal numbers][1]，并每行显示10个.我也必须使用getPentagonalNumber(int n)方法.显然这就是为什么它在那里.

I have to find the first N [pentagonal numbers][1] from 1-100 and display them 10 per line. I have to use the getPentagonalNumber(int n) method as well; that is obviously why it is there.

到目前为止，这是我的代码.

Here is my code so far.

package chapter_5;

public class Five_One {

    public static void main(String[] args) {
        int n = 0;
        int numPerLine = 10;
        for ( n = 0;  n < 11;  n ++)           
    }

    public static int getPentagonalNumber(int n) {
        int formula = n * (3 * n - 1) / 2;
        while ( formula < )
    }
}

推荐答案

这还有很长的路要走.让我们用更好的方法将其分解.

This has a long way to go. Let's break it down here with a better approach.

让我们创建一个返回一定数量的五边形数字的方法(我们将使用数组.)这使我们稍后在可能还有额外功劳的情况下也可以使用该方法！

Let's make a method that returns a set number of pentagonal numbers (we'll use an array.) This allows us to use the method later if perhaps there's extra credit too!

我们的签名如下:

class Jason {

    public static void main(String[] args) {
        // don't mix the calc routine and printing...
        int[] pents = getPentagonals(100); // calcs and stores the first 100 values

        final int numPerLine = 10;
        for(int i = 0; i < pents.length; i++) {
            System.out.print(pents[i]);
            System.out.print(" ");
            if(i % numPerLine == numPerLine - 1) System.out.println("");
        }
    }

    static int[] getPentagonals(int n) {
        int[] pents = new int[n];
        // calculate pents
        for(int i = 0; i < pents.length; i++) {
            // calculate the ith pentagonal here (assuming the 0th is first)
            // store it in pent[i]
        }
        return pents;
    }

}

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