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  1. #include <cmath>
  2. #include <iostream>
  3. #include <vector>
  4.  
  5. struct div_with_positive_remainder_t
  6. {
  7.     int quot;
  8.     int rem;
  9. };
  10.  
  11. div_with_positive_remainder_t div_with_positive_remainder(int x, int y)
  12. {
  13.     if (y == 0) {
  14.         throw std::runtime_error("division by zero");
  15.     }
  16.     int r = x % y;
  17.     if (r < 0) {
  18.         r += std::abs(y);
  19.     }
  20.     int q = (x - r) / y;
  21.     return {q, r};
  22. }
  23.  
  24. std::vector<int> to_negbase(int n, int negbase = -2)
  25. {
  26.     std::vector<int> res;
  27.  
  28.     while (n != 0) {
  29.         auto div = div_with_positive_remainder(n, negbase);
  30.         res.push_back(div.rem);
  31.         n = div.quot;
  32.     }
  33.     return res;
  34. }
  35.  
  36. int to_int(const std::vector<int> &digits, int base = -2) {
  37.     int res = 0;
  38.     int power = 1;
  39.     for (auto digit : digits){
  40.         res += power * digit;
  41.         power *= base;
  42.     }
  43.     return res;
  44. }
  45.  
  46. std::vector<int> negative_base(const std::vector<int> &v)
  47. {
  48.     return to_negbase(-to_int(v));
  49. }
  50.  
  51. void print(const std::vector<int>& v, int negbase = -2)
  52. {
  53.     std::cout << to_int(v, negbase) << " = [";
  54.     for (auto e: v) {
  55.         std::cout << e << " ";
  56.     }
  57.     std::cout << "](" << negbase << ")" << std::endl;
  58. }
  59.  
  60.  
  61. int main()
  62. {
  63.     print(to_negbase(2369, -10), -10); // 18449(-10)
  64.     print(to_negbase(1234, -8), -8);   // 16462(-10)
  65.     print(to_negbase(23));
  66.  
  67.     print(negative_base({1, 1, 0, 1, 0, 1, 1})); 
  68. }
  69.  
Success #stdin #stdout 0s 3416KB
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stdin
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Standard input is empty
stdout
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2369 = [9 4 4 8 1 ](-10)
1234 = [2 6 4 6 1 ](-8)
23 = [1 1 0 1 0 1 1 ](-2)
-23 = [1 0 0 1 1 1 ](-2)
https://ideone.com/BPDcvT
language:
C++14 (gcc 8.3)
created:
9 years ago
visibility:
public

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