leetcode: Add largest lcm

math/src/primes.cpp

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+
+#include <cassert>
+
+#include <vector>
+
+std::vector<int> get_primes(int n) {
+  assert(n > 0);
+  std::vector<bool> primes(true, n + 1);
+  for (int i = 2; i < n + 1; ++i) {
+    if (!primes[i]) {
+      continue;
+    }
+    for (int j = i + 1; j < n + 1; ++j) {
+      if (j % i == 0) {
+        primes[j] = false;
+      }
+    }
+  }
+
+  return {};
+}

src/leetcode/0000.largest_lcm/Cargo.toml

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+[package]
+name = "lc-0000-largest-lcm"
+version = "0.1.0"
+edition = "2021"
+publish = false
+
+[dependencies]

src/leetcode/0000.largest_lcm/Makefile

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+
+build: main.cpp
+	g++ -g main.cpp -o main

src/leetcode/0000.largest_lcm/index.md

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+
+# 求能整除1-n的最小正整数 
+
+```text
+输入一个n (1 <= n <= 10000),
+求能整除1-n的最小正整数，即最小公倍数.
+由于数可能比较大，输出结果 mod 987654321.
+
+例子:
+输入: 3
+输出: 6 (6是能整除1, 2, 3的最小正整数)
+```

src/leetcode/0000.largest_lcm/main.cpp

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+// Copyright (c) 2024 Xu Shaohua <[email protected]>. All rights reserved.
+// Use of this source is governed by General Public License that can be found
+// in the LICENSE file.
+
+#include <cassert>
+
+#include <iostream>
+#include <vector>
+
+int solution(int n) {
+  // 解题思路:
+  // 1. 先计算 1..n 之间的所有质数
+  // 2. 创建一个数组, 用于存放每个质因数的最大出现次数
+  // 3. 遍历 1..n 之间所有整数, 进行因式分解, 然后更新最大质因数数组
+  // 4. 将最大质因数数组中的值相乘, 即可得到结果.
+
+  // 得到质数
+  const int max_num = n + 1;
+  std::vector<bool> primes(max_num, true);
+  for (int i = 2; i <= n; ++i) {
+    if (!primes[i]) {
+      continue;
+    }
+    for (int j = i + 1; j <= n; ++j) {
+      if (j % i == 0) {
+        primes[j] = false;
+      }
+    }
+  }
+
+  std::vector<int> max_factors(max_num, 0);
+  // 遍历所有整数
+  for (int num = 2; num <= n; ++num) {
+    // 遍历所有质数
+    int rem = num;
+    for (int j = 2; j <= num; ++j) {
+      if (!primes[j]) {
+        continue;
+      }
+
+      // 确定当前质数作为质因数的次数.
+      int factor_count = 0;
+      while (rem % j == 0) {
+        rem /= j;
+        factor_count += 1;
+      }
+
+      // 更新质因数的最大个数
+      if (factor_count > max_factors[j]) {
+        max_factors[j] = factor_count;
+      }
+
+      // 当前整数 num 已经被除尽了
+      if (rem == 0) {
+        break;
+      }
+    }
+  }
+
+  // 最后, 计算最大质因数的积.
+  int ans = 1;
+  const int kModular = 987654321;
+  for (int num = 2; num <= n; ++num) {
+    if (max_factors[num] > 0) {
+      for (int i = 0; i < max_factors[num]; ++i) {
+        ans = ans * num % kModular;
+      }
+    }
+  }
+
+  return ans;
+}
+
+void check_solution() {
+  assert(solution(3) == 6);
+}
+
+int main() {
+  check_solution();
+  return 0;
+}

src/leetcode/0000.largest_lcm/src/main.rs

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+// Copyright (c) 2024 Xu Shaohua <[email protected]>. All rights reserved.
+// Use of this source is governed by General Public License that can be found
+// in the LICENSE file.
+
+pub fn solution1() {
+    todo!();
+}
+
+pub type SolutionFn = fn();
+
+fn check_solution(_func: SolutionFn) {
+    todo!();
+}
+
+fn main() {
+    check_solution(solution1);
+}
+
+#[cfg(test)]
+mod tests {
+    use super::{check_solution, solution1};
+
+    #[test]
+    fn test_solution1() {
+        check_solution(solution1);
+    }
+}