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|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.

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>

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数。我知道我想创建一个名为total的变量来存储将所有奇数相加的值，但我不知道如

问仅添加奇数
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回答 6

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