Question

我正在寻找一种有效的算法，给出两个正整数n和m，找到互质对(x,y)的数量，1 <= x <= n和1 <= y <= m }。有什么想法吗？

Answer 1

为什么不使用三元树来生成互质？像

这样的东西

def gen_coprimes(max_m, max_n):
    t1 = [(2,1)]
    t2 = [(3,1)]
    b1 = lambda m,n:(2*m-n, m)
    b2 = lambda m,n:(2*m+n, m)
    b3 = lambda m,n:(m+2*n, n)
    while t1 or t2:
        yield from t1
        yield from t2
        t1 = [(i,j) for i,j in (f(m,n) for m,n in t1 for f in (b1,b2,b3)) if i<max_m and j<max_n]
        t2 = [(i,j) for i,j in (f(m,n) for m,n in t2 for f in (b1,b2,b3)) if i<max_m and j<max_n]

list(gen_coprimes(5,6))提供[(2, 1), (3, 1), (3, 2), (4, 1), (4, 3)]。

应该在O（n）附近。如果我错了，请纠正我，仍然可以处理这个重要的O符号。算法来自：http://en.wikipedia.org/wiki/Coprime_integers#Generating_all_coprime_pairs

给定段中的互质对数

1 个答案: