HDOJ--Least Common Multiple

最新推荐文章于 2019-04-17 20:49:11 发布

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最新推荐文章于 2019-04-17 20:49:11 发布

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分类专栏： OJ 文章标签： HDOJ gcd lcm

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本文介绍了一种求解多个正整数最小公倍数的方法，通过迭代计算两两之间的最小公倍数来得到最终结果，并提供了一个C++实现示例。

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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

思路：最小公倍数=数的乘积除以最大公约数，求一组数的最小公倍数，就迭代求两两的最小公倍数。

坑：数据大会溢出

#include <iostream>
#include <vector>
using namespace std;

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


int main()
{
    int n;
    cin >> n;
    for (int ii = 0; ii < n; ++ii)
    {
        int count = 0;
        cin >> count;
        int result = 1;
        int a = 1;
        for (int i = 0; i < count; ++i)
        {
            int b;
            cin >> b;
            result = a / gcd(a, b)*b;       //怕溢出，所以先除gcd再乘
            a = result;
        }
        cout << result << endl;

    }
    return 0;
}