RosettaCodeData/Task/Zumkeller-numbers/C++/zumkeller-numbers.cpp

144 lines
3.9 KiB
C++

#include <iostream">
#include <cmath>
#include <vector>
#include <algorithm>
#include <iomanip>
#include <numeric>
using namespace std;
// Returns n in binary right justified with length passed and padded with zeroes
const uint* binary(uint n, uint length);
// Returns the sum of the binary ordered subset of rank r.
// Adapted from Sympy implementation.
uint sum_subset_unrank_bin(const vector<uint>& d, uint r);
vector<uint> factors(uint x);
bool isPrime(uint number);
bool isZum(uint n);
ostream& operator<<(ostream& os, const vector<uint>& zumz) {
for (uint i = 0; i < zumz.size(); i++) {
if (i % 10 == 0)
os << endl;
os << setw(10) << zumz[i] << ' ';
}
return os;
}
int main() {
cout << "First 220 Zumkeller numbers:" << endl;
vector<uint> zumz;
for (uint n = 2; zumz.size() < 220; n++)
if (isZum(n))
zumz.push_back(n);
cout << zumz << endl << endl;
cout << "First 40 odd Zumkeller numbers:" << endl;
vector<uint> zumz2;
for (uint n = 2; zumz2.size() < 40; n++)
if (n % 2 && isZum(n))
zumz2.push_back(n);
cout << zumz2 << endl << endl;
cout << "First 40 odd Zumkeller numbers not ending in 5:" << endl;
vector<uint> zumz3;
for (uint n = 2; zumz3.size() < 40; n++)
if (n % 2 && (n % 10) != 5 && isZum(n))
zumz3.push_back(n);
cout << zumz3 << endl << endl;
return 0;
}
// Returns n in binary right justified with length passed and padded with zeroes
const uint* binary(uint n, uint length) {
uint* bin = new uint[length]; // array to hold result
fill(bin, bin + length, 0); // fill with zeroes
// convert n to binary and store right justified in bin
for (uint i = 0; n > 0; i++) {
uint rem = n % 2;
n /= 2;
if (rem)
bin[length - 1 - i] = 1;
}
return bin;
}
// Returns the sum of the binary ordered subset of rank r.
// Adapted from Sympy implementation.
uint sum_subset_unrank_bin(const vector<uint>& d, uint r) {
vector<uint> subset;
// convert r to binary array of same size as d
const uint* bits = binary(r, d.size() - 1);
// get binary ordered subset
for (uint i = 0; i < d.size() - 1; i++)
if (bits[i])
subset.push_back(d[i]);
delete[] bits;
return accumulate(subset.begin(), subset.end(), 0u);
}
vector<uint> factors(uint x) {
vector<uint> result;
// this will loop from 1 to int(sqrt(x))
for (uint i = 1; i * i <= x; i++) {
// Check if i divides x without leaving a remainder
if (x % i == 0) {
result.push_back(i);
if (x / i != i)
result.push_back(x / i);
}
}
// return the sorted factors of x
sort(result.begin(), result.end());
return result;
}
bool isPrime(uint number) {
if (number < 2) return false;
if (number == 2) return true;
if (number % 2 == 0) return false;
for (uint i = 3; i * i <= number; i += 2)
if (number % i == 0) return false;
return true;
}
bool isZum(uint n) {
// if prime it ain't no zum
if (isPrime(n))
return false;
// get sum of divisors
const auto d = factors(n);
uint s = accumulate(d.begin(), d.end(), 0u);
// if sum is odd or sum < 2*n it ain't no zum
if (s % 2 || s < 2 * n)
return false;
// if we get here and n is odd or n has at least 24 divisors it's a zum!
// Valid for even n < 99504. To test n beyond this bound, comment out this condition.
// And wait all day. Thanks to User:Horsth for taking the time to find this bound!
if (n % 2 || d.size() >= 24)
return true;
if (!(s % 2) && d[d.size() - 1] <= s / 2)
for (uint x = 2; (uint) log2(x) < (d.size() - 1); x++) // using log2 prevents overflow
if (sum_subset_unrank_bin(d, x) == s / 2)
return true; // congratulations it's a zum num!!
// if we get here it ain't no zum
return false;
}