hdu 1019 Least Common Multiple(LCM的运用)

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

水题：  求最小公倍数，其实就是两数相乘，然后乘以最大公因数。

           最大公倍数与某个数的最大公倍数还是原本的最大公倍数不会共边

           所以对于这题多个数字来说，我们两个两个算就好了。过程中可能会爆int,所以还是开long long 好点

           那我想到的处理方法就是先接受第一个数字，然后不断输入下一个数，每输入一个数运算一次。

 完整代码如下：#include <iostream>
#include <queue>
#include <string.h>
#include <string>
#include <algorithm>
#include <map>
#include <cstdio>
using namespace std;
long long int zxgbs(long long int a,long long int b)
{
if(a>b)
return zxgbs(b,a);
else
{
if(b%a==0)
return a;
return zxgbs(b%a,a);
}
}
int main()
{
long long int cas,num,a,b,i,j;
// cout<<"s";
cin>>cas;
while(cas--)
{
cin>>num;
cin>>a;
num--;
while(num--)
{
cin>>b;
j=zxgbs(a,b);
//cout<<j<<endl;
a=a*b/j;
}
cout<<a<<endl;
}
return 0;
}