带有return语句的方法不返回任何内容[重复]

Question

我必须写一个方法int SumBeyond（int k）来找到最小的n，使得小于n的自然数之和超过k。但是，当我尝试测试该方法时，它不会返回任何值。

public static int sumBeyond(int k){
    int i=1;
    int sum=0;
    while(i<k){
        sum=sum+i;
        i=i+1;
    }
    return sum;
}

我尝试这样调用函数，在main方法中：

sumBeyond(100);

Answer 1

如上所述，问题不是您的方法没有返回某些内容，而是您没有对返回的内容执行任何操作。此外，如上所述，您专注于寻找总和，但这实际上不是问题所问的。我很赞同您在这里使用while循环，因为它可能根本不会执行，而且您也不知道它会运行多少次。所以我重写了它来检查正确的东西，并改编了deHaar的main来练习它。这让我能够亲自检查答案，因为有些情况下，平等和需要“数字小于”而不是“数字小于或等于”是微妙的。我的数学老师非常喜欢Joop Eggen的二次公式法；这只是很难做到正确（事实上，如果我要做的话，我最终会测试它是否与我这里的一致）。

public class ShadesOfLittleGauss {
    public static int sumBeyond(int k) {
        int i = 1; //have summed (trivially) all natural numbers less than 1 so far
        int sum = 0;
        while (sum <= k) { //needs to exceed, so <=
            sum = sum + i;
            i = i + 1;
        }
        return i;
    }
public static void main(String[] args) {
    for (int k = -1; k < 10; k++) {
        System.out.println("The least number " +
            "such that the sum of the natural numbers smaller than n exceeds " +
            k + " is " + sumBeyond(k));

    }
}
}

输出（您可以手动检查是否正确，如所述）：

The least number such that the sum of the natural numbers smaller than n exceeds -1 is 1
    The least number such that the sum of the natural numbers smaller than n exceeds 0 is 2
    The least number such that the sum of the natural numbers smaller than n exceeds 1 is 3
    The least number such that the sum of the natural numbers smaller than n exceeds 2 is 3
    The least number such that the sum of the natural numbers smaller than n exceeds 3 is 4
    The least number such that the sum of the natural numbers smaller than n exceeds 4 is 4
    The least number such that the sum of the natural numbers smaller than n exceeds 5 is 4
    The least number such that the sum of the natural numbers smaller than n exceeds 6 is 5
    The least number such that the sum of the natural numbers smaller than n exceeds 7 is 5
    The least number such that the sum of the natural numbers smaller than n exceeds 8 is 5
    The least number such that the sum of the natural numbers smaller than n exceeds 9 is 5

更新：我确实解决了二次型，并提出了以下与更简单的方法一致的方法。

public static int sumBeyond2(int k) {
if (k < 0) { //do not take squareroots of negatives
    return 1;
}
double x = -1.0 / 2 + Math.sqrt(1 + 8 * k) / 2; //from solving quadratic inequality n(n+1)/2.0 > k
if (Math.abs(Math.round(x) - x) < 0.0000001) { //special case, it is an integer, so ceil won't reliably add 1
    return 1 + 1 + (int) Math.round(x);
}
return 1 + (int) Math.ceil(x); //adding 1 because of wording, integers less than, ceil because needs to exceed k
}

Answer 2

如果我答对了你的问题，你想找到最小的n，使得小于n的自然数之和超过k，因此，你不应该返回总和本身，因为它不是n，但需要计算才能找到最小的n。

您可以通过以下方式执行此操作：

public static int sumBeyond(int k) {
    int n = 0;
    int sum = 0;

    for (int i = 0; i < k; i++) {
        // provide an intermediate sum (the one before this step) for logging purpose
        int intermediateSum = sum;
        // sum up
        sum += i;
        // set the return value to the current "natural number"
        n = i;
        // print some detailed debug log
        System.out.println("sum:\t" + sum +
                " (" + intermediateSum + " + " + i +  ")\t——>\tn = " + n);

        // exit the loop if the sum is greater than k
        if (sum >= k) {
            break;
        }
    }

    return n + 1;
}

像这样在main中调用它

public static void main(String[] args) {
    int k = 100;
    System.out.println("The least n" +
            + "such that the sum of the natural numbers smaller than n exceeds "
            + k + " is " + sumBeyond(k));
}

将打印

sum:    0 (0 + 0)       ——> n = 0
sum:    1 (0 + 1)       ——> n = 1
sum:    3 (1 + 2)       ——> n = 2
sum:    6 (3 + 3)       ——> n = 3
sum:    10 (6 + 4)      ——> n = 4
sum:    15 (10 + 5)     ——> n = 5
sum:    21 (15 + 6)     ——> n = 6
sum:    28 (21 + 7)     ——> n = 7
sum:    36 (28 + 8)     ——> n = 8
sum:    45 (36 + 9)     ——> n = 9
sum:    55 (45 + 10)    ——> n = 10
sum:    66 (55 + 11)    ——> n = 11
sum:    78 (66 + 12)    ——> n = 12
sum:    91 (78 + 13)    ——> n = 13
sum:    105 (91 + 14)   ——> n = 14
The least n such that the sum of the natural numbers smaller than n exceeds 100 is 15

我真的希望我做对了，还是不确定
哦，如果0不是一个自然数，那么从1开始迭代。

Answer 3

int sum100 = sumBeyond(100);
System.out.println("Sum is " + sum100);

然后小改进：

public static int sumBeyond(int k) {
    int i = 1;
    int sum = 0;
    while (i < k) {
        sum += i;
        ++i;
    }
    return sum;
}

public static int sumBeyond(int k) {
    int sum = 0;
    for (int i = 1; i < k; ++i) {
        sum += i;
    }
    return sum;
}

public static int sumBeyond(int k) {
    // return (k - 1) * (1 + k - 1) / 2;
    return (k - 1) * k / 2;
}

为了解决问题指出：

• 求n使得和等于n-1

所以我们得到：

x² - x - 2k'
------------  >=  0
     2

=0的解决方案：

a = 1/2
b = -1/2
c = -2k'
            _________
    -b +/- V b² - 4ac
x = ------------------
          2a

 x = 1/2 +/- sqrt(1/4 + 4k'/4) =
   = 1/2 +/- 1/2 . sqrt(1 + 4k')

正x

 x = 1/2 + 1/2 . sqrt(4k' + 1)

public static int sumBeyond(int k) {
    double x = (Math.sqrt(4 * (k-1) + 1) + 1) / 2;
    return (int) Math.ceil(x);
}

解决方案应该作为注释给出数学。