biginteger

StackOverflowError computing factorial of a BigInteger?

丶灬走出姿态 提交于 2019-11-26 16:59:51
问题 I am trying to write a Java program to calculate factorial of a large number. It seems BigInteger is not able to hold such a large number. The below is the (straightforward) code I wrote. public static BigInteger getFactorial(BigInteger num) { if (num.intValue() == 0) return BigInteger.valueOf(1); if (num.intValue() == 1) return BigInteger.valueOf(1); return num.multiply(getFactorial(num.subtract(BigInteger.valueOf(1)))); } The maximum number the above program handles in 5022, after that the

BIGINT UNSIGNED value is out of range

[亡魂溺海] 提交于 2019-11-26 16:48:55
问题 I am getting the error BIGINT UNSIGNED value is out of range in '(1301980250 - mydb . news_articles . date )' When I run the query SELECT *, ((1 / log(1301980250 - date)) * 175) as weight FROM news_articles ORDER BY weight; Removing the ORDER BY condition, removes the error too. How can I fix it? Update: The date field contains unix timestamp (ex: 1298944082). The error started appearing after I upgraded MySQL from 5.0.x to 5.5.x Any help please? 回答1: I recently ran into this and found the

C++ handling very large integers

巧了我就是萌 提交于 2019-11-26 16:38:13
I am using the RSA Algorithm for encryption/decryption, and in order to decrypt the files you have to deal with some pretty big values. More specifically, things like P = C^d % n = 62^65 % 133 Now that is really the only calculations that ill be doing. I have tried using Matt McCutchen's BigInteger Library, but I am getting a lot of compiler errors during linking, such as: encryption.o(.text+0x187):encryption.cpp: undefined reference to `BigInteger::BigInteger(int)' encryption.o(.text+0x302):encryption.cpp: undefined reference to `operator<<(std::ostream&, BigInteger const&)' encryption.o(

What does BigInteger having no limit mean?

大城市里の小女人 提交于 2019-11-26 16:36:48
I looked into this stackoverflow question relating to Big Integer and specifically I do not understand this line (the words in italics): In the BigInteger class, I have no limits and there are some helpful functions there but it is pretty depressing to convert your beautiful code to work with the BigInteger class, specially when primitive operators don't work there and you must use functions from this class. I don't know what I am missing but to represent something that has no limit you would require infinite memory ? Whats is the trick here ? There is no theoretical limit. The BigInteger

Javascript summing large integers

安稳与你 提交于 2019-11-26 16:35:50
In javascript I would like to create the binary hash of a large boolean array (54 elements) with the following method: function bhash(arr) { for (var i = 0, L = arr.length, sum = 0; i < L; sum += Math.pow(2,i)*arr[i++]); return sum; } In short: it creates the smallest integer to store an array of booleans in. Now my problem is that javascript apparently uses floats as default. The maximum number I have to create is 2^54-1 but once javascript reaches 2^53 it starts doing weird things: 9007199254740992+1 = 9007199254740994 Is there any way of using integers instead of floats in javascript? Or

`xrange(2**100)` -> OverflowError: long int too large to convert to int

安稳与你 提交于 2019-11-26 15:28:29
xrange function doesn't work for large integers: >>> N = 10**100 >>> xrange(N) Traceback (most recent call last): ... OverflowError: long int too large to convert to int >>> xrange(N, N+10) Traceback (most recent call last): ... OverflowError: long int too large to convert to int Python 3.x: >>> N = 10**100 >>> r = range(N) >>> r = range(N, N+10) >>> len(r) 10 Is there a backport of py3k builtin range() function for Python 2.x? Edit I'm looking for a complete implementation of "lazy" range() , not just a partial implementation of some of its functionality. Okay, here's a go at a fuller

BigInteger equivalent in Swift?

烈酒焚心 提交于 2019-11-26 14:41:13
Is there an equivalent to Java's BigInteger class in Swift? I am tying to do large calculations in Swift with positive integers larger than UInt64's maximum number. What is the best way to handle these numbers? iOS-Coder I'm also working on a BigNumber library with which you can do some big number calculations. Actually the library is based on the GNU Multiple Precision (GMP) library and I wrote an Objective-C / Swift wrapper. Currently big integer mathematics, including a lot of operator overloading, is possible. A code example goes like: var err : NSError? var bi1 = BigInt(nr:

Logarithm for BigInteger

时光总嘲笑我的痴心妄想 提交于 2019-11-26 14:27:00
问题 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? 回答1: 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

The best cross platform (portable) arbitrary precision math library [closed]

若如初见. 提交于 2019-11-26 12:48:51
I'm looking for a good arbitrary precision math library in C or C++. Could you please give me some advices / suggestions? The primary requirements: It MUST handle arbitrarily big integers (my primary interest is on integers). In case that you don't know what the word arbitrarily big means, imagine something like 100000! (the factorial of 100000). The precision MUST NOT NEED to be specified during library initialization / object creation. The precision should ONLY be constrained by the available resources of the system. It SHOULD utilize the full power of the platform, and should handle "small"

How do I convert a String to a BigInteger?

微笑、不失礼 提交于 2019-11-26 12:45:16
问题 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