pow c

Idea One:x,1 x,2 x,4 x,8 x,16 in turn to add, and then down 5=1+2+2, this method is not goodIdea two: Using 5=101=4+1= (2,2) + (2,0), very ingenious Doubleres =1.0; BOOLFlag =false; unsignedintNewn =n;//This step is more magical, in order to take care of the smallest negative number if(n0) {newn=-n;flag=true;} while(newn) {if(Newn 0x01) {res*=x; } newn>>=1; X*=x; } if(flag)return 1/Res; returnResThinking three: Half, recursive thinking, when I can have the idea of recursi

Implement Pow (x, n).Problem Solving Ideas:Directly using the multiplication implementation, note that if n is large, the number of recursion will be too many, so in n=10 and N=-10 place a checkpoint, Java implementation is as follows:static public double Mypow (double x, int n) { if (n==1) return x; else if (n>1nJava for Leetcode 050 Pow (x, N)

Source of the topic https://leetcode.com/problems/powx-n/Implement Pow (x, n). Test instructions Analysis Input:x,nOutput:pow (X,n)Conditions: To meet certain memory requirements, the algorithm is less complex Topic ideas Just started to think that the direct ride, but burst memory, look carefully found in fact can be used by the dichotomy of the information of each multiplication, notice that n can be negative, so the pr

This article mainly introduces how to use the pow () method in JavaScript. it is the basic knowledge in JS learning. if you need it, you can refer to this method to return the base number's exponential power, that is, baseexponent. Syntax Math.pow(base, exponent ) ; The following is the detailed information about the parameters: Base: base number Exponents: Index of the base number Return value: Returns the exponent power of the base number.

Leetcode question No. 372 (Super Pow) Your task is to calculate AB MoD 1337 where A is a positive integer and B are an extremely large positive integer given in The form of an array. Example1:A = 2b = [3]Result:8 Example2:A = 2b = [1,0]result:1024 This question and the 50th question are actually very similar. One of the only notable places 3 The range of integers that the int variable can represent is exc

Implement POW (x, N).NoticeYou don ' t need the precision of your answer, it's acceptable if the expected answer and your answer ' s Diffe Rence is smaller than 1e-3 .ExamplePow(2.1, 3) = 9.261Pow(0, 1) = 0Pow(1, 0) = 1ChallengeO (LOGN) time Public classSolution {/** * @paramx the base number *@paramn the Power number *@returnThe result*/ Public DoubleMypow (DoubleXintN) {//Write Your code here if(n = = 0) return1; if

Title Description:Implement POW (x, N).Ideas:If n is an even number:Mypow (x,n) = Mypow (X,N/2) squaredIf n is an odd number:Mypow (X,n) = x * MYPOW (x, (n-1)/2) squaredImplementation code:public class Solution {public double Mypow (double x, int n) { if (n Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.Leetcode--Pow (x, N)

problem:Implement Pow (x, n).Analysis:This problem inherently are very easy and simple . Don' t try to treat a problem as difficult problem, even the mysterious power operation in Mathmatics, could is so easily and elegnatly sovled through algorithm.soltuion1. The instant idea was to multiple x one by one until N times. The time complexity of such solution is O (n), which is unacceptable!Naive: Public classSolution { Public DoubleMypow (DoubleXintN) {

Topic:Implement Pow (x, n). (Medium)Analysis:The implementation of the library function exponentiation operation, traversal is timed out, with a fast power, is divided into the idea of treatment, each time the n minus half.Note: The value range of n, n = min_int,-n will be over range, here WA once.Code:1 classSolution {2 Private:3 DoubleHelperDoubleXLong LongN) {4 if(n = =0) {5 return 1;6 }7 Doubleresult =1;8

One Day Together Leetcode series (i) Title Implement POW (x, N). (ii) Problem solvingThe topic is simple, to achieve the X-th square.*/* There are a few points to note: When the return value is 12.x==1, when the return value is 1;x==-1, the 3.n==-2147483648, special case, the range of int is determined according to the parity of N 1.n==0 when the -2147483648~2147483647, */classSolution { Public:DoubleMypow (DoubleXintN) {if(n==0|| x=

/** 50. Pow (x, N) * 2016-5-7 by Mingyang * Divide and conquer divided into sub-problems------"recursive * complexity is log (n), as he is D Ividing n by half all the. * I started writing my own code sucks, because I didn't do it completely divede into half, because I kept repeating that process. * The improved code overcomes this problem*/ Public Static DoubleMyPow1 (DoubleXintN) {Doubleres=0.0; if(n==0) return1.0; Booleanflag=f

Implement Pow (x, n).Analysis:X^n = x^ (N/2) *x^ (N/2) (*x, if n is odd).Note:we need consider nSolution:1 Public classSolution {2 Public DoublePowDoubleXintN) {3 if(x==0)return0;4 if(n==0)return1;5 6 DoubleRes;7 if(n>0)8res =powrecur (x,n);9 Else{Ten if(n>integer.min_value) Oneres = Powrecur (x,-n); A Else { -res = Powrecur (x,-(n+1)); -res = res*x; the } -res = 1

The POW function is x*x...*x, where y x multiplies (x is double, y is int type)Consider the situation:1. Because in Java when y=0x=0 is the result of either 1, the corresponding processing is done here;2. When Y is negative, the result should be 1/result;3 Considering the results of a large case; Public DoublePowDoubleXintN) {if(x = = 0.0 N = = 0) { return1; } Doubleres = 1.0; intK =N; if(N ) {k= -N; } while(k! = 0)

Title Description:Implement your own POW (double x, int n) methodImplementation ideas:Consider bit operations. Consider the binary representation of N, take n=51 (110011) as an example, x^51 = x^1*x^2*x^16*x^32, so each time n is unsigned right one bit, and X takes the square of the current value, if n is shifted to the bottomis 1, it will be res*x. Consider the special case, when N is Integer.min_value, at this time-n=integer.max_value+1, I was the f

Tags: leetcode interview questions binary algorithm Encryption My writing method for this question may be different from that of others, and the overall thinking is the same. The first step is to determine N, as shown in the code. I want to convert exponential n to binary, for example, 3 ^ 45. The binary value of 45 is 101101. Scan from left to right and skip the first place. If the bit is 0, the result is squared, that is, the result is multiplied by itself. If the bit is 1, we need to multiply

Implement POW (X,N).Https://oj.leetcode.com/problems/powx-n/Train of Thought: Do not multiply n consecutive times, it will time out. Recursive solution. Do not write it as repeat calculation. There are positive and negative numbers for processing.Public class solution {public double POW (Double X, int N) {If (n = 0) return 1; double res = 1; long limit = N; limit = Limit

Implement POW (X,N), Which calculatesXRaised to the powerN(Xn ). Example 1: Input: 2.00000, 10Output: 1024.00000 Example 2: Input: 2.10000, 3Output: 9.26100 Example 3: Input: 2.00000, -2Output: 0.25000Explanation: 2-2 = 1/22 = 1/4 = 0.25 Note: -100.0 X NIs a 32-bit signed integer, within the range [? 231,231? 1] Maybe I think too much, because the negative number range of the int is more than the positive number range. to convert it to a positive

/** Calculates the integer number of times of the value. CPP * * Created on:2018 April 13 * Author:soyo*/#include#include#includeusing namespacestd;intMain () {DoublePowerDouble Base,intexp); intx=2, y=3; Long intZ; Z=pow (2,3); cout"values are:"Endl; Z=z>>2;//move right Two bitscout"values are:"Endl; DoubleA; clock_t Start,stop,consume_time; Start=clock (); A=power ( A, -); Stop=clock (); Consume_time=stop-start; cout"the new values are:"Endl; cou

Question 1: exclusive or exchange of two numbers Suppose X = 2; y = 3. Our goal is (no tooth decay !!) Exchange the values of X and Y; Take advantage of exclusive or particularity: x ^ x = 0 That is, the two identical numbers are 0; Solution: Make x = x ^ y = 2 ^ 3 Make y = x ^ y = 2 ^ 3 ^ 3 = 2 (x = x ^ y) Make x = x ^ y ^ 2 = 2 ^ 3 ^ 2 = 3 (x = x ^ y; y = 2) Question 2: Implement POW (int x, int y), that is, Power Y of X The yth power of X is that

A simple math on modular calculations.Code below can be simplified of course.classSolution {Const intMOD =1337; Long Long_pow (intAintp) {if(p = =0)return 1; if(p = =1)returnA; Long LongPR = _pow (A, p/2); Long Longr = p%2? A:1; return(PR * PR)% MOD) * R%MOD; } Public: intSuperpow (intA, vectorint>b) {intn =b.size (); if(!n)returnAMOD; VectorLong Long>tmp (n); Long LongCB =A; for(inti = n-1; I >=0; I--) {Tmp[i]= _pow (CB, B[i])%MOD; CB= _pow (CB,Ten) %MOD; } Long LongRET =1; for(inti