通過平均方法傳遞數組，導致百分比低

Question

由於某些原因，當我將數組傳遞給方法時，平均值被填充錯誤，我獲得了非常低的百分比。 幾乎可以看出，由於Array shotsMade只記錄制作鏡頭的整數，而不是錯過它不計算正確的基礎。

import java.util.*;

public class Test {
public static void main(String[] args) {
    int myGameCounter = 1;  
    int shotCount = 0;
    int shotCount1 = 0;
    int [] shotsMade = new int [5];
    int sum = 0;

    System.out.print("Enter Player's Free Throw Percentage: ");
    Scanner input = new Scanner(System.in);
    int percent = input.nextInt();


    //Game #1
    System.out.println("Game " + myGameCounter + ":");
    Random r = new Random();
    myGameCounter++;
    shotCount = 0;
    for (int i = 0; i < 10; ++i){
        boolean in = tryFreeThrow(percent);
        if (in) {
        shotCount++;
        System.out.print("In" + " ");
        }
        else {
        System.out.print("Out" + " ");
        }
    }

    System.out.println("");
    System.out.println("Free throws made: " + shotCount + " out of 10");
    shotsMade[0]= shotCount;

        //Game #2
    System.out.println("");
    System.out.println("Game" + myGameCounter + ":");
    myGameCounter++;
    shotCount1 = 0;
    for (int i = 0; i < 10; ++i){
        boolean in = tryFreeThrow(percent);
        if (in) {
        shotCount1++;
        System.out.print("In" + " ");
        }
        else {
        System.out.print("Out" + " ");
        }   
    }
    System.out.println("");
    System.out.println("Free throws made: " + shotCount1 + " out of 10");
    shotsMade[1]= shotCount1;

    System.out.println("");
    System.out.println("Summary:");



    System.out.println("Best game: " + max(shotsMade)); 
    System.out.println("Total Free Throws Made: " + sum(shotsMade) + " " + "out of 20");
    System.out.println("Average Free Throw Percentage: " + average(shotsMade) +"%");    

}//主要

public static boolean tryFreeThrow(int percent) {
    Random r = new Random();
    int number = r.nextInt(100);
    if (number > percent){ 
     return false;
    }
    return true;
}

public static float average(int nums[]) {
    int total = 0;
    for (int i=0; i<nums.length; i++) {
        total = total + nums[i];
    }
    float f = (total / nums.length);
    return (float)total /(float)nums.length;
}

public static int sum(int nums[]) {
    int sum = 0;
    for (int i=0; i<nums.length; ++i) {
        sum += nums[i];
    }
    return (int)sum;
}

public static int max(int nums[]) {
    int max=nums[0];
    for (int i=1; i<nums.length; i++) {
        if (nums[i] > max) 
            max = nums[i];
    }
    return max;
}

}//類

Answer 1

舊問題，您使用整數算術total / nums.length並返回一個int值。 您稍后將其分配給float ，但該值已被截斷。

只需將其中一個值更改為在除法之前 float ，即vg ((float) total) / num

Answer 2

您正在計算5個號碼的avarage但只設置了2所以，如果所有的出手打你的陣列將是這樣的： 10, 10, 0, 0, 0和avarage將是4。

Answer 3

其中，你的表達

float f = (total / nums.length);

會產生不准確的結果。

total和nums.length都是整數，整數之間的任何操作總是產生整數。

示例：如果total = 10且nums.length = 3，則您希望結果為3.333 ...但實際上結果只是3.只有在此之后才將其轉換為浮點數，結果為3.0。

要獲得所需的結果，您需要在分割之前將兩個整數轉換為浮點數：

float f = (float) total / (float) nums.length;