是否 tom osi std sub while script miss ostream

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 56704 Accepted Submission(s): 21613

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 找出最大那個數，遍歷看是否可以被所有元素整除。不行加上那個最大的數繼續

 1 #include <iostream>
 2 using namespace std;
 3 int main()
 4 {
 5     int i;
 6     int a[1000];
 7     int T;
 8     cin>>T;
 9     while(T--)
10     {
11         int n;
12         int max_number=0;
13         cin>>n;
14         for(i=0;i!=n;++i)
15         {
16             cin>>a[i];
 
17             max_number=a[i]>max_number? a[i]:max_number;
18         }
19         int temp=max_number;
20         while(1)
21         {
22             for(i=0;i!=n;++i)
23             {
24                 if(max_number%a[i]==0)
25                     continue;
26                 else
27                 {
 
28                     max_number+=temp;
29                     break;
30                 }
31             }
32             if(i==n)  break;
33         }
34         cout<<max_number<<endl;
35     }
36 
37     return 0;
38 }

HDU1019 （一組數據的最小公倍數）