Least Common Multiple HDU 1019(一列数的最小公倍数)

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

23 5 7 156 4 10296 936 1287 792 1

Sample Output

10510296

题意： 就是求一列数的最小公倍数；

思路：就是求前两个数的最大公倍数a，再让a与第三个数求最大公倍数再赋予a，这是时候的a就是，前三个数的最大公倍数，依次类推；

在求a，b的最大公倍数时，有一个小技巧，避免溢出  a/gcd(a,b)*b  先除再乘；避免数的溢出；

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int gcd(int a,int b){return b==0?a:gcd(b,a%b);}int main(){int i,j,t,n;scanf("%d",&t);while(t--){scanf("%d",&n);int a=1,b;for(i=0;i<n;i++){scanf("%d",&b);a = a/(gcd(a,b))*b; // 先除再乘，避免数的溢出； }printf("%d\n",a);}return 0;}