问题 A: Least Common Multiple_in arithmetic, the least common multiple (lcm) of -优快云博客

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该程序实例展示了如何利用欧几里得算法计算一组正整数的最小公倍数（LCM）。通过逐个与当前计算的LCM求最小公倍数，避免了溢出问题，保证了在32位整数范围内的计算准确性。样例输入和输出分别验证了这种方法的有效性。

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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.

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：求一组数的最小公倍数，先求第一个数a和第二个数b的最小公倍数M，再将这个得到的数M与第三个数c继续求最小公倍数，直到求完最后一个数字即可。注意求M时x先除以最大公约数D再乘y，防止x乘y时发生溢出。

#include <cstdio>

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

int main() {
    int m, n, x, y;
    scanf("%d", &m);
    while (m--) {
        scanf("%d", &n);
        scanf("%d", &x); //先输入第一个数字
        for (int i = 1; i < n; i++) {
            scanf("%d", &y);
            //x先除以最大公约数再乘y是为了防止溢出
            x = x / gcd(x, y) * y; //将x更新为x与y的最小公倍数
        }
        printf("%d\n", x);
    }
    return 0;
}