算法笔记练习 5.2 最大公约数与最小公倍数 问题 A: Least Common Multiple

算法笔记练习 题解合集

题目链接

题目

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.

输入
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.

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

样例输入

2
2 3 5
3 4 6 12

样例输出

15
12

思路

要求 m 个数的 LCM，先求前两个数的 LCM，再用结果去算和第三个数的 LCM，直到把所有数都算完，结果即为所求。

步骤：

1. 令 $i=1$；
2. 令 $ans=LCM(n_i, n_{i+1})$；
3. $i=i+1$，重复 2 直到 $i+1==m$；
4. 输出 $ans$.

细节：特判一下只有一个数的情况。

代码

#include <stdio.h>

// 返回 a，b 的最大公约数
long gcd(long a, long b){
	return !b ? a : gcd(b, a % b); 
} 

int main() {
	long m, n, i, ans, input;
	while (scanf("%ld", &m) != EOF) {
		while (m--) {
			scanf("%ld", &n);
			scanf("%ld", &ans);
			if (n == 1){
				printf("%ld\n", ans);
				continue; 
			} 
			for (i = 0; i < n-1; ++i) {
				scanf("%ld", &input);
				ans = ans / gcd(ans, input) * input; //ans = LCM(ans,input)
			}
			printf("%ld\n", ans); 
		} 
	} 
	return 0;
}