Least Common Multiple
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 53016 Accepted Submission(s): 20171
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.
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.
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.
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
Source
分析：多个数的最小公倍数，直接每次对两个进行lcm操作就好了，具体代码如下：
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
inline ll gcd(ll a,ll b)
{
return b==?a:gcd(b,a%b);
}
inline ll lcm(ll a,ll b)
{
return a*b/gcd(a,b);
}
ll a[];
int main()
{
ll n;
while(scanf("%lld",&n)!=EOF)
{
while(n--)
{
ll m;
scanf("%lld",&m);
for(ll i=;i<=m;i++)
scanf("%lld",&a[i]);
for(ll i=;i<=m-;i++)
a[i+]=lcm(a[i],a[i+]);
printf("%lld\n",a[m]);
}
}
return ;
}