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.
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
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：
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 inline ll gcd(ll a,ll b)
5 {
6 return b==0?a:gcd(b,a%b);
7 }
8 inline ll lcm(ll a,ll b)
9 {
10 return a*b/gcd(a,b);
11 }
12 ll a[100010];
13 int main()
14 {
15 ll n;
16 while(scanf("%lld",&n)!=EOF)
17 {
18 while(n--)
19 {
20 ll m;
21 scanf("%lld",&m);
22 for(ll i=1;i<=m;i++)
23 scanf("%lld",&a[i]);
24 for(ll i=1;i<=m-1;i++)
25 a[i+1]=lcm(a[i],a[i+1]);
26 printf("%lld\n",a[m]);
27 }
28 }
29 return 0;
30 }
