太简单了。。。题目都不想贴了

 

1 //算n个数的最小公倍数 2 #include
     
       3 #include
      
        4 #include
       
         5 using namespace std; 6 int gcd(int a, int b) 7 { 8     return b==0?a:gcd(b,a%b); 9 }10 int lcm(int a, int b)11 {12     return a/gcd(a,b)*b;13 }14 int main()15 {16     int T;17     scanf("%d",&T);18     while(T--)19     {20         int n;21         scanf("%d",&n);22         int tm = 1;23         int a;24         for(int i = 0; i < n; i++){25             scanf("%d",&a);26             tm = lcm(tm,a);27         }28         printf("%d\n",tm);29     }30     return 0;31 }
       
      
     

 

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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

 

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