biginteger

Converting from Integer, to BigInteger

由 北战南征 提交于 2019-11-27 10:12:08
问题 I was wondering if there was any way to convert a variable of type Integer, to BigInteger. I tried typecasting the Integer variable, but i get an error that says inconvertible type. 回答1: The method you want is BigInteger#valueOf(long val). E.g., BigInteger bi = BigInteger.valueOf(myInteger.intValue()); Making a String first is unnecessary and undesired. 回答2: You can do in this way: Integer i = 1; new BigInteger("" + i); 来源： https://stackoverflow.com/questions/3878192/converting-from-integer

Django BigInteger auto-increment field as primary key?

由 戏子无情 提交于 2019-11-27 09:47:45
问题 I'm currently building a project which involves a lot of collective intelligence. Every user visiting the web site gets created a unique profile and their data is later used to calculate best matches for themselves and other users. By default, Django creates an INT(11) id field to handle models primary keys. I'm concerned with this being overflown very quickly (i.e. ~2.4b devices visiting the page without prior cookie set up). How can I change it to be represented as BIGINT in MySQL and long(

BigInteger: count the number of decimal digits in a scalable method

由 强颜欢笑 提交于 2019-11-27 09:06:07
I need the count the number of decimal digits of a BigInteger . For example: 99 returns 2 1234 returns 4 9999 returns 4 12345678901234567890 returns 20 I need to do this for a BigInteger with 184948 decimal digits and more . How can I do this fast and scalable? The convert-to-String approach is slow: public String getWritableNumber(BigInteger number) { // Takes over 30 seconds for 184948 decimal digits return "10^" + (number.toString().length() - 1); } This loop-devide-by-ten approach is even slower: public String getWritableNumber(BigInteger number) { int digitSize = 0; while (!number.equals

Logarithm for BigInteger

由 可紊 提交于 2019-11-27 09:01:43
I have a BigInteger number, for example beyond 2 64 . Now i want to calculate the logarithm of that BigInteger number, but the method BigInteger.log() does not exist. How do I calculate the (natural) logarithm of my large BigInteger value? If you want to support arbitrarily big integers, it's not safe to just do Math.log(bigInteger.doubleValue()); because this would fail if the argument exceeds the double range (about 2^1024 or 10^308, i.e. more than 300 decimal digits ). Here's my own class that provides the methods double logBigInteger(BigInteger val); double logBigDecimal(BigDecimal val);

Is there a numpy biginteger?

由 萝らか妹 提交于 2019-11-27 07:39:50
问题 Hmm. There doesn't seem to me a way to store Python's bigintegers in a numpy array. Is there something special you have to do, to declare a numpy array with bigints? 回答1: Not specifically, no. You can create an array with dtype='object' , which creates an array of Python objects (including but not limited to ints). This will get you a lot of Numpy array-like functionality but few to none of the performance benefits. Which is to say, an array of Python objects is not significantly different

Large integers in javascript (more the 2^53-1)

由 雨燕双飞 提交于 2019-11-27 07:13:50
问题 What is general principals to operate with large integers in javascript? Like in libraries for bigint? How i do it by myself? 回答1: You may take a look at this implementation. You may also find other implementations useful. 回答2: Another option I've used in the past, is to hand off those operations to a calculation server via jsonp. If you're working with such large numbers, you likely want the improved performance and precision this can give you. 来源： https://stackoverflow.com/questions/3579153

Arbitrarily large integers in C#

由 你说的曾经没有我的故事 提交于 2019-11-27 06:51:54
问题 How can I implement this python code in c#? Python code: print(str(int(str("e60f553e42aa44aebf1d6723b0be7541"), 16))) Result: 305802052421002911840647389720929531201 But in c# I have problems with big digits. Can you help me? I've got different results in python and c#. Where can be mistake? 回答1: Primitive types (such as Int32 , Int64 ) have a finite length that it's not enough for such big number. For example: Data type Maximum positive value Int32 2,147,483,647 UInt32 4,294,967,295 Int64 9

How to convert an 18 digit numeric string to BigInteger?

由 折月煮酒 提交于 2019-11-27 06:48:24
问题 Could anyone help me in converting an 18 digit string numeric to BigInteger in java ie;a string "0x9999999999999999" should appear as 0x9999999999999999 numeric value. 回答1: You can specify the base in BigInteger constructor. BigInteger bi = new BigInteger("9999999999999999", 16); String s = bi.toString(16); 回答2: If the String always starts with "0x" and is hexadecimal: String str = "0x9999999999999999"; BigInteger number = new BigInteger(str.substring(2)); better, check if it starts with "0x"

Long vs BigInteger

由 走远了吗. 提交于 2019-11-27 05:38:58
问题 I understand that both java.lang.Long and java.math.BigInteger can hold very large natural numbers. I also know Long's max value, but what is the max value for BigInteger? And aside from capacity, would BigInteger ever perform better when working with generally large integers that still fall in Long's range? Question Is the only consideration: is my value too large for Long? 回答1: BigInteger is capable of holding far bigger numbers than Long. BigInteger seems capable of holding (2 ^ 32) ^

How do I convert a String to a BigInteger?

由 混江龙づ霸主 提交于 2019-11-27 04:34:47
I'm trying to read some really big numbers from standard input and add them together. However, to add to BigInteger, I need to use BigInteger.valueOf(long); : private BigInteger sum = BigInteger.valueOf(0); private void sum(String newNumber) { // BigInteger is immutable, reassign the variable: sum = sum.add(BigInteger.valueOf(Long.parseLong(newNumber))); } That works fine, but as the BigInteger.valueOf() only takes a long , I cannot add numbers greater than long 's max value (9223372036854775807). Whenever I try to add 9223372036854775808 or more, I get a NumberFormatException (which is
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