HDU 1019 （多个数的最小公倍数）

时间：2021-12-13 21:45:41

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
 
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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==)

        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=,cnt=;

        int x;

        for(int i=;i<n;i++)

        {

            scanf("%d",&x);

            cnt=a/gcd(a,x)*x;//先除后乘，避免溢出

            a=cnt;

        }

        printf("%d\n",a);

    }

    return ;

}

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