九度OJ 1056--最大公约数 1439--Least Common Multiple 【辗转相除法】

题目地址：http://ac.jobdu.com/problem.php?pid=1056

题目描述：

输入两个正整数，求其最大公约数。

输入：

测试数据有多组，每组输入两个正整数。

输出：

对于每组输入,请输出其最大公约数。

样例输入：

49 14

样例输出：

7

来源：

2011年哈尔滨工业大学计算机研究生机试真题

#include <stdio.h>

int gcd1 (int a, int b){
	if (b == 0)
		return a;
	else
		return gcd1 (b, a % b);
}

int gcd2 (int a, int b){
	int tmp;

	while (b != 0){
		tmp = a;
		a = b;
		b = tmp % b;
	}
	return a;
}

int main (void){
	int a, b;

	while (scanf ("%d%d", &a, &b) != EOF){
		printf ("%d\n", gcd1 (a, b));
		printf ("%d\n", gcd2 (a, b));
	}

	return 0;
}

题目地址：http://ac.jobdu.com/problem.php?pid=1439

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
3 5 7 15
6 4 10296 936 1287 792 1

样例输出：

105
10296

#include <stdio.h>
 
int Gcd (int a, int b){
    int tmp;
 
    while (b != 0){
        tmp = a;
        a = b;
        b = tmp % b;
    }
    return a;
}
 
int main(void){
    int n;
    int a;
    int b;
    int m;
 
    while (scanf ("%d", &n) != EOF){
        while (n-- != 0){
            scanf ("%d%d", &m, &a);
            --m;
            while (m != 0){
                scanf ("%d", &b);
                a = a / Gcd (a, b) * b;
                --m;
            }
            printf ("%d\n", a);
        }
    }
 
    return 0;
}

参考资料： 维基百科 -- 辗转相除法