blocks|key|403292|text|teaParty(20,+6)+→+2+1+++X+++++++
teaParty(12,+6)+→+2+1+++X++|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|403293|这两个函数都作为1触发，因为它们都不等于5。它们是由|unstyled|403294|if(tea+!=+5+&&+candy+!=+5)
+++++return+1;|403295|至于|403296|teaParty(5,+9)+→+1++2+++X+++|403297|这是因为它通过了返回1的“五项检查”，并且糖果*2(+10)大于茶(+9)，所以它命中|403298|+if(candy+*+2+>=+tea)
+++++return+2;|403299|然而，您在这里并没有真正要问的问题。你的目标是什么？你能提供更多的上下文吗？也许我们可以用它来找出真正的错误，而不仅仅是症状。|403300|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|V|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$B|C]]|$1|M|3|N|5|F|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$B|C]]|$1|Q|3|R|5|F|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|F|7|12|8|@]|9|@]|A|$]]]|T|$]]

<pre><code>teaParty(20, 6) → 2 1 X 
teaParty(12, 6) → 2 1 X 
</code></pre>

These two are firing as 1 because neither equals 5. They are triggered by

<pre><code>if(tea != 5 &amp;&amp; candy != 5)
 return 1;
</code></pre>

As for

<pre><code>teaParty(5, 9) → 1 2 X 
</code></pre>

That is firing because it passes the "five checks" that return 1, and candy * 2 (which is 10) is greater than tea (which is 9), so it hits

<pre><code> if(candy * 2 &gt;= tea)
 return 2;
</code></pre>

However, you don't really have a question to ask here. What is your objective? Can you provide more context? Perhaps we can use that to find the true error, not just the symptoms.

blocks|key|618982|text|我可以解释为什么你会得到你想要的输出，但我不知道你想让它是什么。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|618983|if(tea+!=+5+&&+candy+!=+5)
+return+1;|code-block|syntax|javascript|618984|这个if语句意味着，如果茶和糖果都不是5，那么返回1。这是您拥有的第三个if语句，因此您需要传递这两个2|618985|if(tea+<+5+%7C%7C+candy+<+5)
+++++return+0;

++if(tea+==+5+&&+candy+==+5)
+++++return+1;|618986|因此，如果茶和糖果都大于或等于5，则第一次失败，如果茶或糖果都不是5，则第二次失败。|618987|在这些情况下|618988|teaParty(20,+6)+→+2+1+++X+++++++
teaParty(12,+6)+→+2+1+++X+
teaParty(10,+20)+→+2+1+++X++++++|618989|茶=+20，12，10都大于5，不是5，糖果=+6，20都大于5，不是5，所以你得到了第三个，如果|618990|这个案例|618991|teaParty(5,+9)+→+1++2+++X++|618992|茶=5糖果=9|618993|所以你得到了第四个如果|618994|++if(candy+*+2+>=+tea)+(9*2+>=+5)+true
+++++return+2;|618995|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|16|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|18|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|19|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|1A|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|1B|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1C|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|1D|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1E|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|1F|8|@]|9|@]|A|$]]|$1|10|3|11|5|D|7|1G|8|@]|9|@]|A|$E|F]]|$1|12|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|13|$]]

I can explain why you are getting the output you are but I don't know what you want it to be.

<pre><code>if(tea != 5 &amp;&amp; candy != 5)
 return 1;
</code></pre>

This if statement means that if both tea and candy are not 5 then return 1. This is the 3rd if statement you have so to get here you need to pass these 2

<pre><code>if(tea &lt; 5 || candy &lt; 5)
 return 0;

 if(tea == 5 &amp;&amp; candy == 5)
 return 1;
</code></pre>

So if tea and candy are both greater than or equal to 5 it will fail the first one and if either tea or candy isn't 5 it will fail the second.

in these cases

<pre><code>teaParty(20, 6) → 2 1 X 
teaParty(12, 6) → 2 1 X 
teaParty(10, 20) → 2 1 X 
</code></pre>

tea = 20, 12, 10 all greater than 5 and not 5
candy = 6, 20 all greater than 5 and not 5
so you get the 3rd if

this case

<pre><code>teaParty(5, 9) → 1 2 X 
</code></pre>

tea = 5
candy = 9

so you get the 4th if 

<pre><code> if(candy * 2 &gt;= tea) (9*2 &gt;= 5) true
 return 2;
</code></pre>

blocks|key|403381|text|if(tea+!=+5+&&+candy+!=+5)在所有测试之前执行，以检查其中一个是否(至少)是另一个的双倍。要么删除该if语句，要么移动它，这样您就可以进行双重检查了。类似于：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|403382|if(tea+<+5+%7C%7C+candy+<+5)
++++return+0;

if(tea+==+5+&&+candy+==+5)
++++return+1;

if(tea+!=+5+&&+candy+!=+5)
{
++++if((candy+*+2+>=+tea)+%7C%7C+(tea+*+2+>=+candy))
+++++++++return+2;

++++return+1;
}|code-block|syntax|javascript|403383|entityMap^0|0|Q|1Q|2|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

<code>if(tea != 5 &amp;&amp; candy != 5)</code> is executed before all your tests to check if one is the double (at least) of the other. Either remove that <code>if</code> statement or move it so you can get your double checks to work. Something like :

<pre><code>if(tea &lt; 5 || candy &lt; 5)
 return 0;

if(tea == 5 &amp;&amp; candy == 5)
 return 1;

if(tea != 5 &amp;&amp; candy != 5)
{
 if((candy * 2 &gt;= tea) || (tea * 2 &gt;= candy))
 return 2;

 return 1;
}
</code></pre>

blocks|key|403399|text|你有一些多余的和不正确的案例。下面是一个更简单的版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|403400|public+int+teaParty(int+tea,+int+candy)+{
++if(tea+<+5+%7C%7C+candy+<+5)++//+your+first+case+was+correct
+++++return+0;

++if(candy+*+2+<=+tea+%7C%7C+tea+*+2+<=+candy)++//+must+be+<=+not+>=
+++++return+2;

++return+1;++//+after+you+handle+the+first+two+cases,+just+return+1
}|code-block|syntax|javascript|403401|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|$]]

You have some redundant and incorrect cases. Here is a simpler version:

<pre><code>public int teaParty(int tea, int candy) {
 if(tea &lt; 5 || candy &lt; 5) // your first case was correct
 return 0;

 if(candy * 2 &lt;= tea || tea * 2 &lt;= candy) // must be &lt;= not &gt;=
 return 2;

 return 1; // after you handle the first two cases, just return 1
}
</code></pre>

blocks|key|619142|text|如果您想要一个不使用Ifs的简单紧凑的单行解决方案，并且在您的代码中检查的一些条件是多余的，这可能会导致错误，则以下是代码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|619143|+public+int+teaParty(int+tea,+int+candy)
+{
++++return+(tea+<+5+%7C%7C+candy+<+5)+?+0:+(tea+>=+2+*+candy+%7C%7C+candy+>=+2+*+tea)+?+2+:+1;
+}|code-block|syntax|javascript|619144|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|$]]

Below is the code if you would like a simple compact one line solution without using Ifs and in your code some conditions checked are redundant which might be causing the errors

<pre><code> public int teaParty(int tea, int candy)
 {
 return (tea &lt; 5 || candy &lt; 5) ? 0: (tea &gt;= 2 * candy || candy &gt;= 2 * tea) ? 2 : 1;
 }
</code></pre>

blocks|key|619215|text|public+int+teaParty(int+tea,+int+candy)+{

++if+(tea>=5+&&+candy>=5)+{
++++if+(tea>=candy*2+%7C%7C+candy>=tea*2)+{+return+2;+}
++++else+{+return+1;+}
++}
++return+0;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|619216|希望这能有所帮助。更清晰的逻辑。|unstyled|619217|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>public int teaParty(int tea, int candy) {

 if (tea&gt;=5 &amp;&amp; candy&gt;=5) {
 if (tea&gt;=candy*2 || candy&gt;=tea*2) { return 2; }
 else { return 1; }
 }
 return 0;
}
</code></pre>

Hope this helps. Clearer logic.

blocks|key|619221|text|public+int+teaParty(int+tea,+int+candy)+{
++if+(tea+>=+5+&&+candy+>=5)+{
++++if+(tea+/+candy+>=+2+%7C%7C+candy+/+tea+>=+2)+{
++++++return+2;
++++}
++++return+1;
++}
++return+0;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|619222|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>public int teaParty(int tea, int candy) {
 if (tea &gt;= 5 &amp;&amp; candy &gt;=5) {
 if (tea / candy &gt;= 2 || candy / tea &gt;= 2) {
 return 2;
 }
 return 1;
 }
 return 0;
}
</code></pre>

I am obviously missing an "if" statement but I am not sure how to cover if tea or candy is not double of each other: 

<a href="http://codingbat.com/prob/p177181" rel="nofollow">Java > Logic-1 > teaParty</a>

We are having a party with amounts of tea and candy. Return the int outcome of the party encoded as 0=bad, 1=good, or 2=great. A party is good (1) if both tea and candy are at least 5. However, if either tea or candy is at least double the amount of the other one, the party is great (2). However, in all cases, if either tea or candy is less than 5, the party is always bad (0). 

<pre><code>teaParty(6, 8) → 1
teaParty(3, 8) → 0
teaParty(20, 6) → 2
</code></pre>

Here is the code I have so far:

<pre><code>public int teaParty(int tea, int candy) {

 if(tea &lt; 5 || candy &lt; 5)
 return 0;

 if(tea == 5 &amp;&amp; candy == 5)
 return 1;

 if(tea != 5 &amp;&amp; candy != 5)
 return 1;

 if(candy * 2 &gt;= tea)
 return 2;

 if(tea * 2 &gt;= candy)
 return 2;


return 0;
}
</code></pre>

These are the errors I am receiving:

<pre><code>Expected This Run 
teaParty(6, 8) → 1 1 OK 
teaParty(3, 8) → 0 0 OK 
teaParty(20, 6) → 2 1 X 
teaParty(12, 6) → 2 1 X 
teaParty(11, 6) → 1 1 OK 
teaParty(11, 4) → 0 0 OK 
teaParty(4, 5) → 0 0 OK 
teaParty(5, 5) → 1 1 OK 
teaParty(6, 6) → 1 1 OK 
teaParty(5, 10) → 2 2 OK 
teaParty(5, 9) → 1 2 X 
teaParty(10, 4) → 0 0 OK 
teaParty(10, 20) → 2 1 X 
other tests
X 
</code></pre>

Java CodingBat Logic teaParty?

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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我显然漏掉了一个" if“语句，但我不确定如果茶或糖果不是彼此的两倍该如何覆盖：我们正在开一个有很多茶和糖果的聚会。返回被编码为0=bad、1=good或2=great的参与方的int结果。如果茶和糖果都至少是5，那么聚会就很棒。但是，如果茶和糖果中的一种至少是另一种的两倍，那么聚会就很棒(2)。然而，在所有情况下，如...

问Java CodingBat逻辑teaParty？
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问Java CodingBat逻辑teaParty？
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