Question

我正在进行硬币翻转任务，并且我的大部分功能正常运行（虽然与我在此处看到的代码相比，它的方式不太优雅）。

我正试图找到一种方法来告诉用户哪个号码在他们的翻转中出现最多，并且如果将头部分配给偶数#s，并将尾部分配给奇数#s，哪一个最多。我正在寻找实施这些功能的建议。

到目前为止，这是代码：

import java.io.*;
import java.util.*;


public class coinFlip {

    public static void main(String[] args) throws IOException {

        // declare in as a BufferedReader; used to gain input from the user
        BufferedReader in;
        in = new BufferedReader(new InputStreamReader(System.in));

        //declare variables
        int flips;
        int anArray[];
        int x;
        int r;
        int counter1 = 0;
        int counter2 = 0;
        int counter3 = 0;
        int counter4 = 0;
        int counter5 = 0;
        int counter6 = 0;
        int counter7 = 0;
        int counter8 = 0;
        int counter9 = 0;
        int counter10 = 0;


            System.out.println("How many times would you like to flip your coin?");
            flips = Integer.parseInt(in.readLine());

            if (flips > 1000) {
            System.out.println("Invalid input, restart program!");
            }

            if(flips <= 1000) {
                System.out.println("You want to flip " + flips + " times");
                anArray = new int[flips];

                for(x = 0; x < flips; x++) {
                    r = (int) Math.round(Math.random()*9)+1;
                    anArray[x] = r;
                    System.out.println(anArray[x]);

                    if (anArray[x] == 1) {
                    counter1 += 1;
                    }
                    else if (anArray[x] == 2) {
                    counter2 += 1;
                    }
                    else if (anArray[x] == 3) {
                    counter3 += 1;
                    }
                    else if (anArray[x] == 4) {
                    counter4 += 1;
                    }
                    else if (anArray[x] == 5) {
                    counter5 += 1;
                    }
                    else if (anArray[x] == 6) {
                    counter6 += 1;
                    }
                    else if (anArray[x] == 7) {
                    counter7 += 1;
                    }
                    else if (anArray[x] == 8) {
                    counter8 += 1;
                    }
                    else if (anArray[x] == 9) {
                    counter9 += 1;
                    }
                    else if (anArray[x] == 10) {
                    counter10 += 1;
                    }
            }

            System.out.println("\n You rolled 1 " + counter1 + " times.");
            System.out.println("You rolled 2 " + counter2 + " times.");
            System.out.println("You rolled 3 " + counter3 + " times.");
            System.out.println("You rolled 4 " + counter4 + " times.");
            System.out.println("You rolled 5 " + counter5 + " times.");
            System.out.println("You rolled 6 " + counter6 + " times.");
            System.out.println("You rolled 7 " + counter7 + " times.");
            System.out.println("You rolled 8 " + counter8 + " times.");
            System.out.println("You rolled 9 " + counter9 + " times.");
            System.out.println("You rolled 10 " + counter10 + " times.");



        }
    }
 }

Answer 1

import java.io.*;
import java.util.Random;

public class CoinFlip {
    public static void main(final String[] args) throws IOException {

        // declare in as a BufferedReader; used to gain input from the user
        final BufferedReader in = new BufferedReader(new InputStreamReader(System.in));

        //declare variables
        System.out.println("How many times would you like to flip your coin?");
        final int flips = Integer.parseInt(in.readLine());;

        if (flips > 1000) {
            System.out.println("Invalid input, restart program!");
            return;
        }

        System.out.println("You want to flip " + flips + " times");
        final int[] counters = new int[10],
                     side     = new int[2];

        int r=0,x,max=0;
        final Random rand = new Random();

        for(x = 0; x < flips; ++x) {
            r = rand.nextInt(10);
            System.out.println(r+1);
            counters[r]++;
        }

        for ( x = 0; x < counters.length; ++x )
        {
            System.out.println("You rolled " + (x+1) + " " + counters[x] + " times.");
            if ( counters[x] > max )
            {
                max = counters[x];
                r = x+1;
            }
            side[x%2] += counters[x];
        }

        System.out.println(r + " was rolled most.");
        System.out.println("You rolled " + side[0] + " heads and " + side[1] + " tails." );
    }
 }

Answer 2

使用循环，如另一个答案所示，将使生活更轻松。

您的代码中存在更严重的逻辑错误：

绕过Math.random()的输出，它给出一个介于0和1之间的浮点数，并将其舍入以得到一个整数。下表显示了您从代码中获得的与RNG相关的输出：

| Math.random() output | Resulting Integer |
| 0    ~ 0.04999       | 0                 |
| 0.05 ~ 0.14999       | 1                 |
| ......               | .....             |
| 0.95 ~ 0.99999       | 10                |

看到问题？ 0和10只出现其他数字的一半。

您应该floor()输出，或使用Random.nextInt()生成统一的整体分布。

Answer 3

这会让你的生活更轻松：

int counter[10];

...

for(x = 0; x < flips; x++) {
    r = (int) Math.round(Math.random()*9)+1;
    anArray[x] = r;
    System.out.println(anArray[x]);
    counter[r-1]++;
}

for(i=1; i <= counter.length; i++)
    System.out.println("You rolled " + i + " " + counter[i-1] + " times.");

对于你的其他两个问题，没有什么可以真正“建议”。只需迭代counter，记住最大值，并分别加上偶数和奇数索引。

Java硬币翻转建议

3 个答案: