largenumber

问题 I have an integer literal in the format 0x75f17d6b3588f843b13dea7c9c324e51 . Is there a way to avoid the compiler syntax error "integer literal is too large to be represented in any integer type"? Because I know I can work with those kinds of types (I'm using uint128_t from the EOS library and if I manually insert it, it works). Is there a way to somehow parse this string directly into the exact same integer at run time? 回答1: 128 bit integer literals are not mandated by the standard, so it's

问题 I have an integer literal in the format 0x75f17d6b3588f843b13dea7c9c324e51 . Is there a way to avoid the compiler syntax error "integer literal is too large to be represented in any integer type"? Because I know I can work with those kinds of types (I'm using uint128_t from the EOS library and if I manually insert it, it works). Is there a way to somehow parse this string directly into the exact same integer at run time? 回答1: 128 bit integer literals are not mandated by the standard, so it's

问题 Input Integers 'n' (up to 10^14) and 'm'(up to 10^3) Output Fib(n) modulo m Sample Cases Input: 239 1000 Output: 161 Input: 2816213588 239 Output: 151 Hint given in Question As it is not possible to iterate 'n' times (because n is huge), consider using Pisano Period(repetition of remainders when every element Fibonacci series is divided by any integer) Code which I wrote (maybe wrong, but passes above-mentioned cases) n, m = map(int, input().split()) a, b = 0, 1 fib_rems = [0, 1] # remainders

问题 I have some large numbers (like 10000000 digits in total or more ). Since I will lose precision, they are stored like strings . Lets take some 'small' numbers as an example (some numbers also can be integers): a = 4782916578987656789087654678908765467417657489104.9486718490129876578 b = 657890876566567890.8765467890987654 I want to make some operations on them. Division . Well, answer is 7.270075858095143e+30 Multiplicaiton: 3.1466371806949598e+66 And so on. Desirable output is full number as

问题 I am trying to find the last digit of sum of Fibonacci Series. I calculate the sum as F(n+2) - 1 . The below code is working fine but it is slow for large numbers (e.g 99999 ). How can I optimize this? n = int(input()) def last_digit(n): a, b = 0, 1 for i in range(n+2): a, b = b, a + b return (a-1) % 10 print(last_digit(n)) 回答1: Look at this table: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibtable.html notice that fib(60) last digit is 0 and fib(61) last digit is 1 , that

问题 I am trying to find the last digit of sum of Fibonacci Series. I calculate the sum as F(n+2) - 1 . The below code is working fine but it is slow for large numbers (e.g 99999 ). How can I optimize this? n = int(input()) def last_digit(n): a, b = 0, 1 for i in range(n+2): a, b = b, a + b return (a-1) % 10 print(last_digit(n)) 回答1: Look at this table: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibtable.html notice that fib(60) last digit is 0 and fib(61) last digit is 1 , that

问题 I have the following equation: c = SHA-256(Q, K(i,0), K(i,1), ..., K(i,k-1), h(i,0), h(i,2), …, h(i,k-1)) mod q and the input values: k1 =

问题 It seems like bash's maximum signed integer value is 9223372036854775807 (2^63)-1. Is there a way for bash to handle larger values than this? I need to handle numbers up to 10000000000000000000000000001, but I'm not sure how to accomplish this in bash. A=10000000000000000000000000000 echo $A 10000000000000000000000000000 let A+=1 echo $A 4477988020393345025 EDIT Thanks Benjamin W. for your comment. Based on that I am trying the following strategy. Are there any perceived issues with this?

问题 This question already has answers here : Counting trailing zeros of numbers resulted from factorial (10 answers) Closed last month . I'm trying to count the number of trailing zero with a factorial. e.g 4! = 24 So you retrieve 0. 9! = 362880 So you retrieve 1. 10! = 9! x 10 = 3628800 So you retrieve 2. 11! = 10! x 11 = 3.99168E7 So you retrieve 2. static double factorial(double n) { double f = 1; for(int i = 1 ; i <= n ; i++) { f *= i; } return f; } static int numberOfZeros(double f) { int

问题 I have a dataset on which I'm trying to apply some arithmetical method. The thing is it gives me relatively large numbers, and when I do it with numpy, they're stocked as 0. The weird thing is, when I compute the numbers appart, they have an int value, they only become zeros when I compute them using numpy. x = np.array([18,30,31,31,15]) 10*150**x[0]/x[0] Out[1]:36298069767006890 vector = 10*150**x/x vector Out[2]: array([0, 0, 0, 0, 0]) I have off course checked their types: type(10*150**x[0