Java（奇数和计算平均值）

我想不出这个

//Modify the following to sum only the odd numbers from 1 to 100, and compute the average. (Hint: n is an odd number if n % 2 is not 0.)
    sum=0;
    for (int number = 1; number <= 100; ++number) { // for loop
         sum += number;     // same as "sum = sum + number"
    }
    System.out.println(sum);

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你需要测试这个数字是否是奇数，并计算你找到的赔率

int oddCount = 0;
int sum = 0;
for (int number = 1; number <= 100; ++number) { // for loop
  if (number % 2 != 0) {
     sum += number;
     oddCount++;
  }
}
System.out.println(sum / (double) oddCount)

或者，跳过偶数值，例如

int oddCount = 0;
int sum = 0;
for (int number = 1; number <= 100; number += 2) { // <-- skip the even values.
  sum += number;
  oddCount++;
}
System.out.println(sum / (double) oddCount)

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假设奇数除以2时余数为1，则不需要if语句：

int oddCount = 0, sum = 0;
for (int n = 1; n <= 100; ++n) { // for loop
    oddCount += n % 2;
    sum += n * (n % 2);
}
System.out.println(sum / (double) oddCount)

---

这里的诀窍是，偶数的n%2为零，它们对总数没有影响，但奇数的n%2为1，因此它们才有意义。

注释告诉您该怎么做。