问题 A: Least Common Multiple

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;
}

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