hdu_1019Least Common Multiple(最小公倍数)

太简单了。。。题目都不想贴了

 //算n个数的最小公倍数
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 int gcd(int a, int b)
 {
     return b==?a:gcd(b,a%b);
 }
 int lcm(int a, int b)
 {
     return a/gcd(a,b)*b;
 }
 int main()
 {
     int T;
     scanf("%d",&T);
     while(T--)
     {
         int n;
         scanf("%d",&n);
         int tm = ;
         int a;
         for(int i = ; i < n; i++){
             scanf("%d",&a);
             tm = lcm(tm,a);
         }
         printf("%d\n",tm);
     }
     return ;
 }

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Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45603    Accepted Submission(s): 17131

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

 

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