杭电 1019 Least Common Multiple（最小公倍数）

Least Common Multiple

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

Total Submission(s): 29574    Accepted Submission(s): 11176

Problem Description

The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7 and 15 is 105.

 

 

Input

Input will consist of multiple problem instances. The first line of the input will contain a single integer indicating the number of problem instances. Each instance will consist of a single line of the form m n1 n2 n3 ... nm where m is the number of integers in the set and n1 ... nm are the integers. All integers will be positive and lie within the range of a 32-bit integer.

 

 

Output

For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.

 

 

Sample Input

2 3 5 7 15 6 4 10296 936 1287 792 1

 

 

Sample Output

105 10296

 

 

Source

East Central North America 2003, Practice

 题意：

给定几组数据，判断给定数据的最小公倍数

难点：

最小公倍数的 求解公式

思路：

最小公倍数的求解公式为

两个数的乘积，与两个数的最大公约数的比值。

体会：

这道题 共提交了五次 前三次超时，改了一下第四次wa,我彻底无语了 最后一看 忘了将看最后结果的while（1）；给删了 诶。。。。。还是要细心啊

代码如下：

#include
#include
int gcd(int a,int b)
{
    int t;
    if(a<=b)
    t=a,a=b,b=t;
    while(b)
    {
            t=b;
            b=a%b;
            a=t;
            }
    return a;
    }

//{return !b?a:gcd(b,a%b);}
int main()
{
    int n,a,m;
    scanf("%d",&n);
    while(n--)
    {
              long long sum;
              scanf("%d%d",&m,&a);
              m-=1;
              sum=a;
              while(m--)
              {
                        scanf("%d",&a);
                        sum=sum*a/gcd(sum,a);
                        }
                        printf("%lld\n",sum);
              }
    return 0;
    }