5N的多次方

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N=eval(input(" 请输入一个数:")) for i in range (5): print(pow(N,i))…

Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array. Example1: a = 2 b = [3] Result: 8 Example2: a = 2 b = [1,0] Result: 1024 Credits:Special thanks to @Stomac…

[LeetCode] Power of Four 判断4的次方数

Given an integer (signed 32 bits), write a function to check whether it is a power of 4. Example: Given num = 16, return true. Given num = 5, return false. Follow up: Could you solve it without loops/recursion? Credits:Special thanks to @yukuairoy fo…

[LeetCode] Power of Three 判断3的次方数

Given an integer, write a function to determine if it is a power of three. Follow up:Could you do it without using any loop / recursion? Credits:Special thanks to @dietpepsi for adding this problem and creating all test cases. 这道题让我们判断一个数是不是3的次方数,在Le…

[LeetCode] Power of Two 判断2的次方数

Given an integer, write a function to determine if it is a power of two. Hint: Could you solve it in O(1) time and using O(1) space? 这道题让我们判断一个数是否为2的次方数,而且要求时间和空间复杂度都为常数,那么对于这种玩数字的题,我们应该首先考虑位操作 Bit Operation.在LeetCode中,位操作的题有很多,比如比如Repeated DNA Seque…

[LeetCode] Pow(x, n) 求x的n次方

Implement pow(x, n). 这道题让我们求x的n次方,如果我们只是简单的用个for循环让x乘以自己n次的话,未免也把LeetCode上的想的太简单了,一句话形容图样图森破啊.OJ因超时无法通过,所以我们需要优化我们的算法,使其在更有效的算出结果来.我们可以用递归来折半计算,每次把n缩小一半,这样n最终会缩小到0,任何数的0次方都为1,这时候我们再往回乘,如果此时n是偶数,直接把上次递归得到的值算个平方返回即可,如果是奇数,则还需要乘上个x的值.还有一点需要引起我们的注意的是n有可能…

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