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.
InputInput 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.
OutputFor 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 这道题挺简单的,但我居然想着去分解质因数然后乘一遍,又果断的TLE了.但其实真的很简单,不断地读入一个数,然后求它和之前所有数的lcm,这个怎么求呢?除去两数的gcd即可.说起来还是太菜了.... 代码如下:
#include<cstdio>
using namespace std; long long ans=,n,i,x; long long gcd(long long a,long long b)
{
if(b>a)
{
long long w=a;a=b;b=w;
}
if(a%b==)
{
return b;
}
else
{
return gcd(b,a%b);
}
} int main()
{
long long t;
scanf("%lld",&t);
while(t--)
{
ans=;
scanf("%lld",&n);
for(i=;i<=n;i++)
{
scanf("%lld",&x);
ans*=x/gcd(ans,x);
}
printf("%lld\n",ans);
}
return ;
}
每日刷题身体棒棒!