java 编写一个函数，它接受 int n 并返回小于 n 的奇数之和
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writing a function that takes int n and returns the sum of odd less than n
提问by Eng.Fouad
Here is my code
这是我的代码
public static int sumOfOddLessThan(int n)
{
int iResult = 0;
for(int i = n - 1; i > 0 && i % 2 != 0; i--)
{
iResult = iResult + i;
}
return iResult;
}
It does not work correctly, I dunno why :\
它不能正常工作，我不知道为什么：\
It should return 4 when I enter 5 but it returns 0
当我输入 5 时它应该返回 4 但它返回 0
回答by Finbarr
Your conditional in the for loop reads:
你在 for 循环中的条件是：
i is greater than 0 and i is not even.
i is greater than 0 and i is not even.
When you call the method with 5 as argument, the first value of i will be 4, which is even and therefore the loop does not get evaluated.
当您使用 5 作为参数调用该方法时， i 的第一个值将是 4，这是偶数，因此不会评估循环。
for(i = n-1; i > 0; i++) {
if(i%2==0) {
iResult += i;
}
}
回答by Argote
You're putting the condition i % 2 != 0in the for loop instead of an ifinside of the loop, hence if it's not met even once it breaks out of the entire loop.
您将条件i % 2 != 0放在 for 循环中而不是循环if内部，因此即使它脱离整个循环也没有满足。
Your code should look like this:
您的代码应如下所示：
public static int sumOfOddLessThan(int n)
{
int iResult = 0;
for(int i = n - 1; i > 0; i--)
{
if(i % 2 != 0) {
iResult = iResult + i;
}
}
return iResult;
}
Then again you don't even need a loop, you can evaluate it directly by getting the number of odd numbers lower than Nand squaring that.
然后，您甚至不需要循环，您可以通过使奇数的数量低于N并对其进行平方来直接评估它。
回答by x.509
you should modify the forumla used for adding the series, all you gotta do is to modify it
您应该修改用于添加系列的论坛，您要做的就是修改它
earlier
早些时候
int i = (n+1)/2;
return (i*i)
modified
修改的
int i = n/2;
return (i*i);
TESTinput 1: return 0;
TEST输入1：返回0；
input 2: return 1;
输入2：返回1；
input 3: return 1;
输入3：返回1；
input 4: return 4;
输入4：返回4；
input 5: return 4;
输入5：返回4；
input 6: return 9;
输入6：返回9；
and so on ..
等等 ..
回答by David Ruttka
The second part of a for loop is a continuation condition. In your case, your continuation condition is i > 0 && i % 2 != 0.
for 循环的第二部分是继续条件。在您的情况下，您的继续条件是i > 0 && i % 2 != 0。
For n = 5, the first i is 4, and 4 % 2 is 0. Your continuation condition is not met, and this is why your for loop exits before it begins.
对于 n = 5，第一个 i 是 4，而 4 % 2 是 0。不满足继续条件，这就是 for 循环在开始之前退出的原因。
Try
尝试
for(int i = n - 1; i > 0; i--)
{
if (i % 2 != 0)
{
iResult = iResult + i;
}
}
回答by OmerGertel
The problem is that when the condition in the for is false, the loop exits.
问题是当 for 中的条件为假时，循环退出。
So for 5, i=4and i % 2 != 0is false, so the loop isn't accessed at all.
所以对于 5, i=4andi % 2 != 0是假的，所以根本没有访问循环。
Try this instead:
试试这个：
for(i=((n-1)%2==0?n-2:n-1 ; i>0; i=i-2)
{
i > 0 && i % 2 != 0;
}
Note that by reducing 2 from iat each step, you don't have to check parity on every loop.
请注意，通过i在每一步减少 2 ，您不必在每个循环中检查奇偶校验。
回答by TimCodes.NET
First you are setting i as n-1, so it would be 4 if n is 5, then your condition on the for loop states that i must be odd, which 4 is not, so it doesn't even do one loop. Try this:
首先，您将 i 设置为 n-1，因此如果 n 为 5，它将是 4，那么您在 for 循环上的条件表明 i 必须是奇数，而 4 不是，因此它甚至不会执行一个循环。试试这个：
public static int sumOfOddLessThan(int n)
{
int iResult = 0;
for(int i = n-1; i > 0; i--)
{
if (i % 2 != 0) iResult += i;
}
return iResult;
}
