zoj 1797 Least Common Multiple

ZOJ Problem Set - 1797

Least Common Multiple

---

Time Limit: 1 Second      Memory Limit: 32768 KB

---

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
Submit    Status

---

Code
//1807615 2009-03-28 09:51:26 Wrong Answer  1797 C++ 0 184 Wpl 
//1807641 2009-03-28 09:59:03 Accepted  1797 C++ 0 184 Wpl 
//注意用long long
#include <iostream>
using namespace std;
long long gcd(long long a,long long b)  //求出最大公约数
{
    long long i;
 if(a==0||b==0)   //如果a和b之中有一个是0,则最大公约数就是0
  return 0;
    if(a<b)   //进行交换将a和b中大的给a小的给b
 {
  i=a;
  a=b;
  b=i;
 }
    while(a%b)  //辗转相除的核心,求最大公约数
    {
        i=a%b;
  a=b;
  b=i;
    }
    return b;  //b就是最大公约数
}
int main()
{
 int t,n,i;
 long long s,b;
 scanf("%d",&t);
 while(t--)
 {
  scanf("%d",&n);
  scanf("%lld",&s);
  for(i=1;i<n;i++)
  {
   scanf("%lld",&b);  //输入一个b
   long long temp=gcd(s,b);
   if(temp==0)  //如果最大公约数是0,那么最小公倍数就是0
    s=0;
   else
    s=s*b/temp;   //最大公约数是temp,那么最小公倍数就是a*b/temp(求a和b的最小公倍数)
  }
  printf("%lld\n",s);
 }
 return 0;
}