数学/素因数分解
(test/math/prime_factorization.test.cpp)

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• Last update: 2023-02-25 01:48:23+09:00
• Problem: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=NTL_1_A

Depends on

• 素因数分解 (prime factorization) (include/emthrm/math/prime_factorization.hpp)

Code

/*
 * @title 数学/素因数分解
 *
 * verification-helper: PROBLEM http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=NTL_1_A
 */

#include <iostream>
#include <utility>
#include <vector>

#include "emthrm/math/prime_factorization.hpp"

int main() {
  int n;
  std::cin >> n;
  const std::vector<std::pair<int, int>> ans = emthrm::prime_factorization(n);
  std::cout << n << ": ";
  for (int i = 0; std::cmp_less(i, ans.size()); ++i) {
    for (int j = 0; j < ans[i].second; ++j) {
      std::cout << ans[i].first
                << " \n"[std::cmp_equal(i + 1, ans.size())
                         && j + 1 == ans[i].second];
    }
  }
  return 0;
}

#line 1 "test/math/prime_factorization.test.cpp"
/*
 * @title 数学/素因数分解
 *
 * verification-helper: PROBLEM http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=NTL_1_A
 */

#include <iostream>
#include <utility>
#include <vector>

#line 1 "include/emthrm/math/prime_factorization.hpp"



#line 6 "include/emthrm/math/prime_factorization.hpp"

namespace emthrm {

template <typename T>
std::vector<std::pair<T, int>> prime_factorization(T n) {
  std::vector<std::pair<T, int>> res;
  for (T i = 2; i * i <= n; ++i) {
    if (n % i == 0) [[unlikely]] {
      int exponent = 0;
      for (; n % i == 0; n /= i) {
        ++exponent;
      }
      res.emplace_back(i, exponent);
    }
  }
  if (n > 1) res.emplace_back(n, 1);
  return res;
}

}  // namespace emthrm


#line 12 "test/math/prime_factorization.test.cpp"

int main() {
  int n;
  std::cin >> n;
  const std::vector<std::pair<int, int>> ans = emthrm::prime_factorization(n);
  std::cout << n << ": ";
  for (int i = 0; std::cmp_less(i, ans.size()); ++i) {
    for (int j = 0; j < ans[i].second; ++j) {
      std::cout << ans[i].first
                << " \n"[std::cmp_equal(i + 1, ans.size())
                         && j + 1 == ans[i].second];
    }
  }
  return 0;
}

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