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.
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
/***********************
简单数学题，求连续n个数的最小公倍数
**************************/
Code:
#include <iostream>
#include<string.h>
using namespace std;
int a[10000];
int gcd(int a,int b)//求最大公约数
{
int temp;
if(a<b)
{
temp = a;a = b;b = temp;
}
return (b==0)?a:gcd(b,a%b);
}
int LCM(int a,int b)//求最小公倍数
{
return a/gcd(a,b)*b; //开始写成 a*b/gcd(a,b),WA了两次，因为可能会溢出
}
int main()
{
int T,n,ans,x;
cin>>T;
while(T--)
{
//memset(a,0,sizeof(a));
ans = 1;
cin>>n>>x;
ans = LCM(ans,x);
for(int i = 0;i<n-1;i++)
{
cin>>x;
ans = LCM(ans,x);
}
cout<<ans<<endl;
} return 0;
}