問題 A: Least Common Multiple

題目描述

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.

輸入

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.

輸出

For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.

樣例輸入

2
2 3 5
3 4 6 12

樣例輸出

15
12

思路：

先求兩個數a,b的最大公約數m，然後 a / m * b 則爲 a和b的最小公倍數，同理，擴拓展到求多個數的最小公倍數。

#include<bits/stdc++.h>
using namespace std;

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

int main()
{
    int t;
    while(cin >> t)
    {
        while(t--)
        {
            int n;
            cin >> n;
            vector<int>res(n);
            for(int i=0; i<n; i++)
            {
                cin >> res[i];
            }
            int ans = res[0];
            for(int i=1; i<n; ++i)
            {
                int temp = gcd(ans, res[i]);
                ans = ans / temp * res[i];
            }
            cout << ans << endl;

        }
    }
    return 0;
}