LeetCode OJ:Pow(x, n) (幂运算)】的更多相关文章

Implement pow(x, n). 幂运算,简单的方法snag然很好实现,直接循环相乘就可以了,但是这里应该不是那种那么简单,我的做法使用到了一点递归: class Solution { public: double myPow(double x, int n) { if(n == INT_MIN) return 1.0/(x * myPow(x, INT_MAX)); //这里考虑了一下由于传参数的限制还是将其变成INT_MAX ) return 1.0/myPow(x, -n); )…
题目链接:https://leetcode.com/problems/powx-n/?tab=Description   Problem:实现幂运算即 pow(x,n)   设形式为pow(x,n) = x^n,则有一下分支: 当x==0时,返回0 当n==0时,返回1 当n<0时,(此时需要注意,不能直接将n = -n,因为最小负数变为相反数之后会超过int的最大范围)      需要判断if( n == Integer.MIN_VALUE) 先对n++  然后n = -n ;  x = 1/…
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一,两种不同的求幂运算 求解x^n(x 的 n 次方) ①使用递归,代码如下: private static long pow(int x, int n){ if(n == 0) return 1; if(n == 1) return x; if(n % 2 == 0) return pow(x * x, n / 2); else return pow(x * x, n / 2) * x; } 分析: 每次递归,使得问题的规模减半.2到6行操作的复杂度为O(1),第7行pow函数里面的x*x操作…
Invoker Problem Description On of Vance's favourite hero is Invoker, Kael. As many people knows Kael can control the elements and combine them to invoke a powerful skill. Vance like Kael very much so he changes the map to make Kael more powerful.  In…
链接:http://poj.org/problem?id=1026 Cipher Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 21436   Accepted: 5891 Description Bob and Alice started to use a brand-new encoding scheme. Surprisingly it is not a Public Key Cryptosystem, but t…
CARDS Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 1448   Accepted: 773 Description Alice and Bob have a set of N cards labelled with numbers 1 ... N (so that no two cards have the same label) and a shuffle machine. We assume that N i…
codevs 2541 幂运算  时间限制: 1 s  空间限制: 128000 KB  题目等级 : 钻石 Diamond 题目描述 Description 从m开始,我们只需要6次运算就可以计算出m31: m2=m×m,m4=m2×m2,m8=m4×m4,m16=m8×m8,m32=m16×m16,m31=m32÷m. 请你找出从m开始,计算mn的最少运算次数.在运算的每一步,都应该是m的正整数次方,换句话说,类似m-3是不允许出现的. 输入描述 Input Description 输入为一…
Problem Link: http://oj.leetcode.com/problems/interleaving-string/ Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2. For example, Given: s1 = "aabcc", s2 = "dbbca", When s3 = "aadbbcbcac", return t…
Problem link: http://oj.leetcode.com/problems/reverse-words-in-a-string/ Given an input string, reverse the string word by word. For example, Given s = "the sky is blue", return "blue is sky the". LeetCode OJ supports Python now! The s…