HDU 1019 最小公倍数

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本文提供了一道经典的算法竞赛题目解析——计算一组正整数的最小公倍数（LCM）。通过示例输入输出说明了如何使用辗转相除法求最大公约数，并进一步计算最小公倍数。代码实现简洁高效，适用于ACM竞赛及算法学习。

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1019

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30988 Accepted Submission(s): 11726

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

求多个数的最小公倍数

最小公倍数=两个数相乘除以最大公约数；

为了防止溢出，先除再乘

#include <iostream>
#include <cstdio>

using namespace std;

int gcd(int a,int b)
{
    if(a%b==0)
        return b;
    return gcd(b,a%b);
}

int main()
{
    int n;
    scanf("%d",&n);
    int a[100000];
    while(n--)
    {
        int m;
        scanf("%d",&m);
        for(int i=0;i<m;i++)
        {
            scanf("%d",&a[i]);
        }
        int temp = a[0];

        for(int i=1;i<m;i++)
        {
            temp = temp/gcd(temp,a[i])*a[i];
        }
        printf("%d\n",temp);
    }
    return 0;
}