HDU 1019 （多个数的最小公倍数）

传送门：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): 61592    Accepted Submission(s): 23486

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

East Central North America 2003, Practice

 

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题目意思：

求n个数的最小公倍数

直接暴力即可：暴力规则：求第一个和第二个元素的最小公倍数，然后拿求得的最小公倍数和第三个元素求最小公倍数，继续下去，直到没有元素

小技巧：

通过最大公约数求最小公倍数的时候，先除再乘，避免溢出

code：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
int gcd(int a,int b)
{
    if (b==0)
        return a;
    return gcd(b, a%b);
}
int main()
{
    //公式：a,b的最小公倍数等于a,b的乘积除以a,b的最大公约数
    //直接暴力即可
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        int a=1,cnt=1;
        int x;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&x);
            cnt=a/gcd(a,x)*x;//先除后乘，避免溢出
            a=cnt;
        }
        printf("%d
",a);
    }
    return 0;
}