Lowest Common Multiple Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 51223 Accepted Submission(s): 21224

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
   
   
    思路：注意用long long
   
   
    代码：
   
   
    
    #include<stdio.h> 
long long gcd(long long a,long long b)
{long long c=a*b;while(b){long long t=a%b;a=b;b=t;}return c/a;
}
int main()
{long long n,a[1000];while(scanf("%d",&n)!=EOF){long long k=0;for(int i=0;i<n;i++){scanf("%lld",&a[i]);}k=a[0];for(int i=1;i<n;i++){k=gcd(k,a[i]);//printf("%d\n",k);}printf("%lld\n",k);}return 0;
}

HDU--2028-Lowest Common Multiple Plus

Lowest Common Multiple Plus