ACM1019:Least Common Multiple
Problem Description
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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两个数的乘积等于这两个数的最大公约数与最小公倍数的积。
#define _CRT_SECURE_NO_WARNINGS #include <stdio.h> int gcd(int a, int b); int main() { int row, N, ans, b; scanf("%d", &row); while (row) { scanf("%d%d", &N, &ans); for (int i = 0; i < N - 1; i++) { scanf("%d", &b); //两个数的乘积等于这两个数的最大公约数与最小公倍数的积,先除再乘防止越界。 ans = ans / gcd(ans, b) * b; } printf("%d\n", ans); row--; } return 0; } int gcd(int a, int b) { return b ? gcd(b, a%b) : a; }