HDOJ 2028 Lowest Common Multiple Plus

Lowest Common Multiple Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 49459 Accepted Submission(s): 20445



Problem Description

求n个数的最小公倍数。

Input

输入包含多个测试实例，每个测试实例的开始是一个正整数n，然后是n个正整数。

Output

为每组测试数据输出它们的最小公倍数，每个测试实例的输出占一行。你可以假设最后的输出是一个32位的整数。

Sample Input

2 4 6
3 2 5 7

Sample Output

12
70

#include <cstdio>
using namespace std;

long long gcd(long long a, long long b)
{
if (b == 0)
return a;
else
gcd(b, a % b);
}

long long lcm(long long a, long long b)
{
return a * b / gcd(a, b);
}

int main()
{
long long n, a, b, res;
while (scanf("%I64d", &n) != EOF){
scanf("%I64d", &a);
for (int i = 1; i < n; i++){
scanf("%I64d", &b);
a = lcm(a, b);
}
printf("%I64d\n", a);
}
return 0;
}