HDU1019

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

首先：两个数的最小公倍数为两个数相乘再除以那两个数的最大公约数

但是！！！！！

num1 = num1 / gcd(num1,num2) * num2;//竟然是这里错了
//写成 num1 =（num1 * num2） / gcd(num1,num2) ;就不对

现在还不知道为啥前者WA,改成后者就A了？？！！

 #include<cstdio>
#include<iostream>
#include<queue>
using namespace std;
//两数的最小公倍数=两数相乘 再除以两数的最大公约数 
int gcd(int a,int b)
{
return b == 0 ? a : gcd(b , a % b);
}
int main()
{
int t,n,num1,num2;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&num1);
for(int i = 1;i < n;i++)
{
scanf("%d",&num2);
num1 = num1 / gcd(num1,num2) * num2;//竟然是这里错了
//写成 num1 =（num1 *  num2） / gcd(num1,num2) ;就不对 
}
printf("%d\n",num1);
}
return 0;
}