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This is a very smart observation:
http://www.martinkysel.com/hackerrank-sherlock-and-gcd-solution/
# http://www.martinkysel.com/hackerrank-sherlock-and-gcd-solution/ # from functools import reduce def gcd(a, b): if a == 1 or b == 1: return 1 if a % b == 0: return b return gcd(b, a % b) def multiple_gcd(arr): return reduce(lambda x,y: gcd(x, y), arr) ################# T = int(input()) for i in range(0,T): n = int(input()) arr = list(map(int, input().split())) if len(arr) < 2: print ("NO") continue if (multiple_gcd(arr) == 1): print ("YES") else: print ("NO")
HackerRank - "Sherlock and GCD"
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原文地址:http://www.cnblogs.com/tonix/p/4331426.html