entityMap|blocks|key|14hrg|text|Java中的int类型可用于表示-2147483648到2147483647之间的任意整数。看起来你已经达到了int类型可以容纳的限制。考虑使用long。因为你使用的方法需要很长的时间。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3r8am|来自：|6qhq0|int+convertedLong+=+(int)+n;|code-block|syntax|javascript|2s0nv|至：|2miql|long+convertedLong+=+n;|704su^0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|O|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|P|9|@]|A|@]|B|$]]|$2|E|4|F|6|G|8|Q|9|@]|A|@]|B|$H|I]]|$2|J|4|K|6|7|8|R|9|@]|A|@]|B|$]]|$2|L|4|M|6|G|8|S|9|@]|A|@]|B|$H|I]]|$2|N|4|-4|6|7|8|T|9|@]|A|@]|B|$]]]]

The int type in Java can be used to represent any whole number from -2147483648 to 2147483647.
It seems you've reach the limitation of what the int type can hold. Consider using long. As the method you are using takes a long.
From:
<pre><code>int convertedLong = (int) n;
</code></pre>
To:
<pre><code>long convertedLong = n;
</code></pre>

entityMap|blocks|key|d2m43|text|问题来了，因为您的convertedLong是int类型。为什么？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3nooc|将其设置为long，并将i变量设置为long。|offset|length|style|CODE|c9nfd|出现此错误是因为您仅通过1000000000调用该函数。|e5loh|将L附加到它，以将其作为"long“传递，如|7c81p|100000000000L|code-block|syntax|javascript|8a116|否则它将被当作int。|akvb6^0|0|C|1|0|C|A|0|1|1|0|0|7|3|0^^$0|$]|1|@$2|3|4|5|6|7|8|U|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|V|9|@$E|W|F|X|G|H]]|A|@]|B|$]]|$2|I|4|J|6|7|8|Y|9|@$E|Z|F|10|G|H]]|A|@]|B|$]]|$2|K|4|L|6|7|8|11|9|@$E|12|F|13|G|H]]|A|@]|B|$]]|$2|M|4|N|6|O|8|14|9|@]|A|@]|B|$P|Q]]|$2|R|4|S|6|7|8|15|9|@$E|16|F|17|G|H]]|A|@]|B|$]]|$2|T|4|-4|6|7|8|18|9|@]|A|@]|B|$]]]]

Issue is coming because your convertedLong is of type int. Why?
Make it long and also variable <code>i</code> to be long.
This error is coming because you are calling the function by just <code>1000000000</code>.
Append <code>L</code> to it, to pass it as &quot;long&quot; like
<pre><code>100000000000L
</code></pre>
else it will take as <code>int</code>.

entityMap|0|type|LINK|mutability|MUTABLE|data|url|https://stackoverflow.com/questions/6834037/initialize-a-long-in-java|blocks|key|8cilj|text|除非另有说明，否则java中的所有数字文字都被定义为int类型。因此，您需要做的是将文本定义为long，方法是向文本添加L后缀。|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|cf7bm|1000000000000L|2ne84|另外，为什么要将long转换为int？想要破坏目的吗？|b6hmg|有关更多信息，请查看this问题|157cr^0|Q|3|1B|4|1O|1|0|0|E|0|8|4|F|3|0|A|4|0|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|T|G|@$H|U|I|V|J|K]|$H|W|I|X|J|K]|$H|Y|I|Z|J|K]]|L|@]|6|$]]|$A|M|C|N|2|E|F|10|G|@$H|11|I|12|J|K]]|L|@]|6|$]]|$A|O|C|P|2|E|F|13|G|@$H|14|I|15|J|K]|$H|16|I|17|J|K]]|L|@]|6|$]]|$A|Q|C|R|2|E|F|18|G|@]|L|@$H|19|I|1A|A|1B]]|6|$]]|$A|S|C|-4|2|E|F|1C|G|@]|L|@]|6|$]]]]

All number literals in java, unless otherwise specified, are typed as an <code>int</code>. So what you'll need to do is define your literal as a <code>long</code> by adding the <code>L</code> suffix to the literal.
<code>1000000000000L</code>
Also why are you casting your <code>long</code> to an <code>int</code>? that seesm to defeat the purpose?
Check out <a href="https://stackoverflow.com/questions/6834037/initialize-a-long-in-java">this</a> question for more info

entityMap|blocks|key|38bn6|text|您正在尝试使用以下参数repeatedString("anyText",1000000000000)调用该方法，对吗？这甚至不能编译，你必须在数字的末尾扩展一个L，让编译器知道这是一个long，而不是int，所以一开始你像这样调用你的函数：repeatedString("anyText",1000000000000L)，但如果你仔细想想它是没用的，因为你把它转换成一个int，然后删除int无法容纳的所有字节。因此，停止将long转换为int，只使用给定的参数作为long。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3fpfn^0|B|13|28|1|2K|4|2S|3|3C|14|55|3|5D|3|5W|4|63|3|6H|4|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@$A|I|B|J|C|D]|$A|K|B|L|C|D]|$A|M|B|N|C|D]|$A|O|B|P|C|D]|$A|Q|B|R|C|D]|$A|S|B|T|C|D]|$A|U|B|V|C|D]|$A|W|B|X|C|D]|$A|Y|B|Z|C|D]|$A|10|B|11|C|D]]|E|@]|F|$]]|$2|G|4|-4|6|7|8|12|9|@]|E|@]|F|$]]]]

You are trying to call the method with the following parameters <code>repeatedString(&quot;anyText&quot;,1000000000000)</code> right? This would not even compile you have to extend a <code>L</code> at the end of your number to give the compiler the information that this is a <code>long</code> and not an <code>int</code>, so at first you call your function like this: <code>repeatedString(&quot;anyText&quot;,1000000000000L)</code> but if you think about it it's useless because you casting it to an <code>int</code> which then removes all the bytes which the <code>int</code> is not able to hold. So stop converting the <code>long</code> to an <code>int</code> and just use the given parameter as <code>long</code>.

entityMap|blocks|key|3iol3|text|试试这个。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|60b0k|static+long+repeatString(String+s,+long+n)+{
++++int+length+=+s.length();
++++long+mod+=+n+%25+length;
++++long+count+=+0,+rest+=+0;
++++for+(int+i+=+0;+i+<+length;+%2B%2Bi)+{
++++++++if+(s.charAt(i)+==+'a')+{
++++++++++++%2B%2Bcount;
++++++++++++if+(i+<+mod)
++++++++++++++++%2B%2Brest;
++++++++}
++++}
++++return+count+*+(n+/+length)+%2B+rest;
}

public+static+void+main(String[]+args)+{
++++String+s+=+"aabca";
++++for+(long+i+=+0;+i+<+10;+%2B%2Bi)
++++++++System.out.println(i+%2B+"+:+"+%2B+repeatString(s,+i));
++++long+x+=+1000000000000L;
++++System.out.println(x+%2B+"+:+"+%2B+repeatString(s,+x));
}|code-block|syntax|javascript|2u6kt|输出：|5hcq7|0+:+0
1+:+1
2+:+2
3+:+2
4+:+2
5+:+3
6+:+4
7+:+5
8+:+5
9+:+5
1000000000000+:+600000000000|9q4cq^0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|M|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|N|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|O|9|@]|A|@]|B|$]]|$2|J|4|K|6|E|8|P|9|@]|A|@]|B|$F|G]]|$2|L|4|-4|6|7|8|Q|9|@]|A|@]|B|$]]]]

Try this.
<pre><code>static long repeatString(String s, long n) {
 int length = s.length();
 long mod = n % length;
 long count = 0, rest = 0;
 for (int i = 0; i &lt; length; ++i) {
 if (s.charAt(i) == 'a') {
 ++count;
 if (i &lt; mod)
 ++rest;
 }
 }
 return count * (n / length) + rest;
}

public static void main(String[] args) {
 String s = &quot;aabca&quot;;
 for (long i = 0; i &lt; 10; ++i)
 System.out.println(i + &quot; : &quot; + repeatString(s, i));
 long x = 1000000000000L;
 System.out.println(x + &quot; : &quot; + repeatString(s, x));
}
</code></pre>
output:
<pre><code>0 : 0
1 : 1
2 : 2
3 : 2
4 : 2
5 : 3
6 : 4
7 : 5
8 : 5
9 : 5
1000000000000 : 600000000000
</code></pre>

The method is the following:
<pre><code>public static long repeatedString(String s, long n) {
 
 
 int convertedLong = (int) n;
 String repeated = new String(new char[convertedLong]).replace(&quot;\0&quot;, s);
 System.out.println(repeated);

 long numberOfAs = 0;
 for (int i = 0; i &lt; convertedLong; i++ )
 {
 if (repeated.charAt(i) == 'a')
 {
 numberOfAs += 1;
 }
 }
 System.out.print(numberOfAs);
 return numberOfAs;
}
</code></pre>
method works fine when the parameter <code>n</code> is small, however when it is large like <code>1000000000000</code> it throws an exception:
<pre><code>The literal 1000000000000 of type int is out of range
</code></pre>

the literal x of type int is out of range

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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 方法如下： public static long repeatedString(String s, long n) {            int convertedLong = (int) n;    String repeated = new String(new char[convertedLong]).replace("\0", s);    System.out.println(re

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