HDU 1019最小公倍数

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 53380 Accepted Submission(s): 20323

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

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思路

1、这题也是险过，用了穷举方法，只要清楚一组数据中的最小公倍数肯定是这组数中最大的数的整数倍

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代码

#include <iostream>
#include <stdio.h>
using namespace std;
int main()
{
    int h;
    cin>>h;
    while(h--)
    {
        int a[100001]={0};
        cin>>a[0];
        int count=1;
        for(int i=1;i<=a[0];i++)
        {
            cin>>a[i];
            count++;
        }
        int max=a[1];
        int min=a[1];
        for(int i=2;i<count;i++)
        {
            if(max<a[i])
              max=a[i];
            if(min>a[i])
              min=a[i];
        }
        int p;
        for( p=max;;p+=max)
        {
            int opp=0;
            for(int i=1;i<count;i++)
            {
                if(p%a[i]!=0)
                {
                    opp++;
                    break;
                }
            }
            if(opp==0)
                break;
        }
        cout<<p<<endl;
    }
    return 0;
}