hdu 1019 Least Common Multiple

心已赠人 2022-06-15 05:53 147阅读 0赞

# Least Common Multiple #

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 52181 Accepted Submission(s): 19805**

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
      
       
      
       
    
      
       
      
       
        
       题目解析：
      
       
      
       
        
       这个题就是让求几个数的最小公倍数，首先我们要对数列进行排序，找到最大值，开始向上增大，每次增大最大的数，用循环就可以找到最小公倍数
      
       
      
       
    
      
       
      
       
        
       代码：
      
       
      
       
       
        #include <stdio.h>
    #include <string.h>
    #include <math.h>
    #include <algorithm>
    using namespace std;
    
    int main()
    {
        int t,s,n,i,j,m,a[5000];
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d",&n);
            for(i=0;i<n;i++)
                scanf("%d",&a[i]);
            sort(a,a+n);
            for(i=a[n-1];;i+=a[n-1])
            {
                for(j=0;j<n-1;j++)
                    if(i%a[j]!=0)
                        break;
                if(j==n-1)
                    break;
            }
            printf("%d\n",i);
        }
        return 0;
    }