Sherlock and Squares

题目链接

写代码：

def squares(a, b): counts = 0 for i in range(a,b+1): s = int(math.sqrt(i)) if pow(s,2) == i: counts += 1 return counts

思路是遍历a到b，将每个元素i开平方根并取整，如果取整后的数字平方后与i相等，则是完全平方数，counts加1. 结果：

报错：Your code did not execute within the time limits.

改后代码：

def squares(a, b): counts = 0 end = round(math.sqrt(b)) for i in range(1,end+1): s = i ** 2 if s in range(a,b+1): counts += 1 return counts

思路是从1开始遍历，若这个数的平方在a和b之间，则counts加1. 也可以从a开平方根后四舍五入后的数开始遍历，只要这个数的平方在a到b之间，counts则加1：

def squares(a, b): counts = 0 start = round(a ** 0.5) end = round(b ** 0.5) for i in range(start,end + 1): if i ** 2 in range(a,b+1): counts += 1 return counts

用while循环语句写：

def squares(a, b): counts = 0 c = 1 d = c ** 2 while d <= b: if d >= a: counts += 1 c += 1 d = c ** 2 return counts