#include <iostream> using namespace std;

// 计算最大公约数 int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }

// 计算最小公倍数 int lcm(int a, int b) { return a / gcd(a, b) * b; // 由于b是2,可以简化为 a * 2 / gcd(a, 2) }

int main() { int n; cin >> n; // 计算n和2的最小公倍数 cout << lcm(n, 2) << endl; return 0; }

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Information

ID
25
Time
1000ms
Memory
256MiB
Difficulty
10
Tags
# Submissions
10
Accepted
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