杭电acm2029-Palindromes _easy version

 1 #include<iostream>
 2 using namespace std;
 3 
 4 long long gcd(long long a, long long b) {
 5     int c = a%b;
 6     if (!c)
 7         return b;
 8     a = b, b = c;
 9     gcd(a, b);
10 }
11 int main() {
12     int n;
13     while ((cin>>n)&&n)
14     {
15         long long num[1000];
16         for (int i = 0;i < n;i++)
17             cin >> num[i];
18         for (int i = 0;i < n - 1;i++) 
19             num[i + 1] = num[i]*num[i+1]/gcd(num[i], num[i + 1]);
20         cout << num[n - 1] << endl;
21     }23     return 0;
24 }