<1> 用辗转相除法求最大公约数

算法描述:

m对n求余为a, 若a不等于0

则 m <- n, n <- a, 继续求余

否则 n 为最大公约数

<2> 最小公倍数 = 两个数的积 / 最大公约数*/-< 1>　with the common denominator Euclidean algorithm descr iption: m for n remainder of a, if a is not equal to 0 m < - n, n < - a, otherwise I continue to ask for the greatest common divisor of n < 2>　the least common multiple of two numbers = product/greatest common divisor* /