Least Common Multiple (HDU - 1019) 【简单数论】【LCM】【欧几里得辗转相除法】

Least Common Multiple (HDU - 1019) 【简单数论】【LCM】【欧几里得辗转相除法】

标签： 入门讲座题解 数论

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题目描述

 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.

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

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题意

给定几个数，求这几个数的最小公倍数。

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解析

这是一道基础的欧几里得辗转相除法的题。可以使用朴素gcd算法不断求最小公倍数。

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通过代码

/*
Problem
    HDU - 1019
Status
	Accepted
Memory
	1372kB
Length
	466
Lang
	G++
Submitted
	2019-11-25 22:13:37
Shared

RemoteRunId
	31637683
*/

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

int gcd(int a, int b)
{
    return b? gcd(b, a % b): a;
}

int lcm(int a, int b)
{
    return a / gcd(a, b) * b;   //注意此处不是 a * b / gcd，这样容易爆int.
}
int main()
{
    int times;
    scanf("%d", &times);

    while(times --){
        int n;
        int t1, t2;

        scanf("%d%d", &n, &t1);

        for(int i = 1; i < n; i ++){
            scanf("%d", &t2);

            t1 = lcm(t1, t2);   //不断地将前面求得的最小公倍数当作a，新输入的数当作b，继续求最小公倍数.
        }

        printf("%d\n", t1);
    }

    return 0;
}

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