bluecode-小U的防御挑战

问题描述

小U最近沉迷于一款冒险游戏，在闯关过程中，他通过击败小兵获得了防御宝石，每个防御宝石的价值为 aiai​。经过一番激战，小U终于遇到了关底的boss，这个boss拥有攻击宝石，每颗攻击宝石的价值为 bibi​。小U想知道他手中的防御宝石是否能够抵抗住boss的攻击。
小U的防御成功条件是：所有防御宝石的乘积 ∏i=1nai∏i=1n​ai​ 可以被所有攻击宝石的乘积 ∏i=1mbi∏i=1m​bi​ 整除。你能帮小U判断他是否能够抵抗住boss的攻击吗？

测试样例

样例1：

输入：n = 2 ,m = 5 ,arrayn = [10, 12] ,arraym = [2, 3, 5, 2, 1]
输出：'yes'

样例2：

输入：n = 4 ,m = 5 ,arrayn = [7, 2, 5, 3] ,arraym = [2, 4, 5, 6, 1]
输出：'no'

样例3：

输入：n = 3 ,m = 4 ,arrayn = [6, 3, 9] ,arraym = [2, 3, 4, 1]
输出：'no'

#include <algorithm>
#include <iostream>
#include <map>
#include <vector>

using namespace std;

vector<int> generatePrimes(int limit) {
  vector<bool> isPrime(limit + 1, true);
  isPrime[0] = isPrime[1] = false;
  for (int i = 2; i * i < limit; ++i) {
    if (isPrime[i]) {
      for (int j = i * i; j <= limit; j += i) {
        isPrime[j] = false;
      }
    }
  }

  vector<int> primes;
  for (int i = 2; i <= limit; ++i) {
    if (isPrime[i]) {
      primes.push_back(i);
    }
  }

  return primes;
}

map<int, int> primeFactors(int x, const vector<int> &primes) {
  map<int, int> factors;
  for (int p : primes) {
    if (p * p > x)
      break;
    while (x % p == 0) {
      factors[p]++;
      x /= p;
    }
  }

  if (x > 1)
    factors[x]++;

  return factors;
}

std::string solution(int n, int m, const std::vector<int> &arrayn,
                     const std::vector<int> &arraym) {
  int max_num = 0;
  for (int num : arraym)
    max_num = max(max_num, num);
  for (int num : arrayn)
    max_num = max(max_num, num);

  vector<int> primes = generatePrimes(max_num);

  map<int, int> A_factors;
  for (int num : arrayn) {
    map<int, int> factors = primeFactors(num, primes);
    for (auto &[p, cnt] : factors) {
      A_factors[p] += cnt;
    }
  }

  map<int, int> B_factors;
  for (int num : arraym) {
    map<int, int> factors = primeFactors(num, primes);
    for (auto &[p, cnt] : factors) {
      B_factors[p] += cnt;
    }
  }

  for (auto &[p, cnt] : B_factors) {
    if (A_factors[p] < cnt) {
      return "no";
    }
  }

  return "yes";
}

int main() {
  // Add your test cases here

  std::cout << (solution(2, 5, {10, 12}, {2, 3, 5, 2, 1}) == "yes")
            << std::endl;
  std::cout << (solution(4, 5, {7, 2, 5, 3}, {2, 4, 5, 6, 1}) == "no")
            << std::endl;

  return 0;
}