blocks|key|5674891|text|import+acm.program.*;

public+class+AddingOddNumbers+extends+ConsoleProgram+{
++++public+void+run()+{
++++++++int+n+=+readInt("enter+a+positive+nunber:+");
++++++++int+total+=+0;
++++++++for+(int+i+=+0;+i+<+n;+i%2B%2B)+{
++++++++++++if+(n+==+1)+{
++++++++++++++++println(1);
++++++++++++}+else+{
++++++++++++++++println((i+*+2)+%2B+1);
++++++++++++++++total+%2B=+(i+*+2)+%2B+1;
++++++++++++}
++++++++}
++++++++println("total+:+"+%2B+total);
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5674892|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>import acm.program.*;

public class AddingOddNumbers extends ConsoleProgram {
 public void run() {
 int n = readInt("enter a positive nunber: ");
 int total = 0;
 for (int i = 0; i &lt; n; i++) {
 if (n == 1) {
 println(1);
 } else {
 println((i * 2) + 1);
 total += (i * 2) + 1;
 }
 }
 println("total : " + total);
 }
}
</code></pre>

blocks|key|3464595|text|sum+=+0;

for+(i+=+1;+i+<+n*2;+i=i%2B2)
++++sum+=+sum+%2B+i;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3464596|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

blocks|key|3464600|text|这将为您提供奇数sum。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3464601|++if+(n>0)+
+++{
++++total=0;
++++++for+(int+i+=+1;+i+<+n;+i+%2B%2B){
++++++++if+(i%252+==+1)
++++++++++++total%2B=i;
++++++}
+++}|code-block|syntax|javascript|3464602|如果希望包含n，则将条件更改为i<=n。|offset|length|style|CODE|3464603|entityMap^0|0|0|F|4|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

This will give you the odd number sum.

<pre><code> if (n&gt;0) 
 {
 total=0;
 for (int i = 1; i &lt; n; i ++){
 if (i%2 == 1)
 total+=i;
 }
 }
</code></pre>

If you want to inclusive of n, then change the condition to <code>i&lt;=n</code>.

blocks|key|3464621|text|我也会对奇数使用循环。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3464622|for+(int+i+=+0,+j+=+1;+i+<+n;+i%2B%2B,+j+%2B=+2)+{
++++println(j);
++++total+%2B=+j;
}
println(total);|code-block|syntax|javascript|3464623|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I would use a loop for the odd numbers as well.

<pre><code>for (int i = 0, j = 1; i &lt; n; i++, j += 2) {
 println(j);
 total += j;
}
println(total);
</code></pre>

blocks|key|116260|text|int+oddSum+=+0;
for+(int+i+=+0;+i+<+n;+i%2B%2B){
++oddSum+=+oddSum+%2B+(i*2)+%2B+1;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|116261|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>int oddSum = 0;
for (int i = 0; i &lt; n; i++){
 oddSum = oddSum + (i*2) + 1;
}
</code></pre>

blocks|key|3947291|text|也许你知道如何计算所有数字的和，直到给定的数字n？公式很简单：(n+*+(n%2B1))/2。现在只得到奇数的和有点棘手，但是-不用担心，你可以只使用前面的公式。首先请注意，直到给定数字n的所有偶数的和是：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3947292|如果N是偶数(即直到n/2乘以2的所有数字的和)，这是因为如果N是奇数，你就有奇数，如果N是奇数，你就会有奇数|3947293|事实上，如果你有整数除法，公式总是：(((n/2)*+(n/2%2B1))/2)+*+2+=+(n/2)*+(n/2%2B1)|3947294|因此，要计算所有奇数的和，直到n，您只需从所有数字的和中减去偶数的和：|3947295|(n+*+(n%2B1))/2+-+(n/2)*(n/2%2B1)|3947296|事实上，如果你仔细观察，你会注意到1%2B3%2B...(2*n-1)总是等于n%5E2。|3947297|这个答案应该可以帮助你解决所有语言的问题，我把代码留给你了。字面上只有一行。|3947298|entityMap^0|N|1|V|D|0|0|I|15|0|0|0|T|0|H|E|Z|3|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Y|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|12|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|N|3|O|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|P|3|Q|5|6|7|1B|8|@]|D|@]|E|$]]|$1|R|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|S|$]]

Maybe you know how to compute the sum of all numbers up to a given number <code>n</code>? The formula is quite simple: <code>(n * (n+1))/2</code>. Now getting the sum of only the odd numbers is a bit trickier but - no worries you can make use only of the previous formula for that. First notice that the sum of all even numbers up to a given number n is:

<ul>
<li><code>(((n/2)* (n/2+1))/2) * 2</code> if N is even(i.e. the sum of all numbers up to n/2 times two that is because you have <code>2+4+6+8+...N = 2*(1+2+3+...n/2)</code>)</li>
<li><code>((((n-1)/2)* ((n-1)/2+1))/2) * 2</code> if N is odd</li>
</ul>

In fact if you have integer division the formula is always: <code>(((n/2)* (n/2+1))/2) * 2 = (n/2)* (n/2+1)</code>

So to compute the sum of all the odd numbers up to n you simply subtract the sum of the even numbers from the sum of the all numbers:

<code>(n * (n+1))/2 - (n/2)*(n/2+1)</code>

In fact if you observe closely you will notice that the sum <code>1+3+...(2*n-1)</code> always equals to <code>n^2</code>.

This answer should help you solve your problem in all languages and I am leaving the code to you. It is literally one line.

So the question I'm trying to solve the user is supposed to enter any positive number. Then I'm trying to write a program that adds only the odd numbers up to the number the user enters and displays the total. So for example if the user enters 4 my program should add four odd numbers. 1 + 3 + 5 + 7 = 16. The only tools I have available are for statement, if, if/else if,while loop and println. 

I can only figure out how to print out the odd numbers. I know I want to create a variable named total to store the value of adding up all the odd numbers but I don't know how that fits into the program. 

<pre><code>import acm.program.*;

public class AddingOddNumbers extends ConsoleProgram {
 public void run() {
 int n = readInt("enter a positive nunber: ");
 int total = 0;
 for (int i = 0; i &lt; n; i++) {
 if (n == 1) {
 println(1);
 } else {
 println((i * 2) + 1);
 }
 }
 }
}
</code></pre>

adding only odd numbers

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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所以我试图解决的问题是，用户应该输入任何正数。然后，我尝试编写一个程序，只将奇数加到用户输入的数字上，并显示总数。例如，如果用户输入4，我的程序应该添加四个奇数。1+3+5+7= 16。我唯一可用的工具是for statement、if、if/else if、while loop和println。我只能想出如何打印出奇...

问仅添加奇数
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