HDU 1019 Least Common Multiple【gcd+lcm+水+多个数的lcm】

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

East Central North America 2003, Practice

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1019

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 }