exam 461

public class Solution { public int HammingDistance(int x, int y) { int distance = 0; string sX = Convert.ToString(x, 2); string sY = Convert.ToString(y, 2); int maxLength = Math.Max(sX.Length, sY.Length); //填充0，使两个字符串右对齐 sX = sX.PadLeft(maxLength, ‘0‘); sY = sY.PadLeft(maxLength, ‘0‘); for ( int i = 0 I MaxLength I ++) { if (sX[i] != sY[i]) { distance++; } } return distance; } } From for notes (Wiz)

();} returnnum*Flag;}intInitintsum) { intans=0; memset (q,0,sizeof(q)); for(intI=1; i) { intx=Xu[i]; if(sum>=x) {ans+=x*p; Q[i]+=x; Sum-=x; X=0; } Else if(sum>0) {ans+=sum*p; Q[i]+=sum; X-=sum; Sum=0; } ////////////Slow Wash for(intj=1; j) if(q[j]>0) { if(q[j]>=x) {ans+=x*v2; Q[J]-=x; Q[i]+=x; X=0; Break; } Else{ans+=q[j]*v2; Q[i]+=Q[j]; X-=Q[j]; Q[J]=0; } } ///////////fast-washing if(!x)Cont

end of the loop, only one number is not 0, the other number is 0, and at this point 0 corresponds to all the binary is 0, at this time as long as the number of the calculation is not 0 of binary 1 of the number.Improvement: The above uses three while, with one is enough:1 classSolution {2 Public:3 intHammingdistance (intXinty) {4 intdis =0;5 while(x | |y) {//or operation, as long as there is a not 0 for this loop6Dis + = (x 1) ^ (Y 1);7X >>=1;8Y >>=1;9 }Ten

The Hamming distance between and integers is the number of positions at which, the corresponding bits is different.Given integers x y and, calculate the Hamming distance.PS: To find the Hamming distance.Idea: The number of x and y binary xor 1Publicclasssolution{publicinthammingdistance (Intx, inty) {//stringx1= Integer.tobinarystring (x); //stringy1= Integer.tobinarystring (y);inttem=x^y; intcount=0;stringstr =integer.tobinarystring (TEM); for (Inti=0;iLeetcode

Problem Description:The Hamming distance between two integers refers to the number of different positions of these two numbers corresponding to bits.Give two integers x and y calculate the Hamming distance between them.Attention:0≤ x , y Example:input: x = 1, y = 4 output: 2 Explanation:1 (0 0 0 1 ) 4 (0 1 0 0) ↑ ↑ The above arrows pointed to the different positions of the corresponding bits.Method: 1 class Solution (object): 2 def Hammingdistance (self, x, y): 3 Span style

The Hamming distance between and integers is the number of positions at which, the corresponding bits is different.Given integers x y and, calculate the Hamming distance.Train of thought: Count two binary bitwise compare the number of different bits, first XOR, then the number of binary 1 of XOR. public class solution { public int hammingdistance (int x, int y) { int xor = x ^ y; int distance = 0; while (xor! = 0 ++; XOR = (xor-1); // " return distance; }}LeetCode-

fibs number of digits problem:461 Time limit:1000ms Memory limit:65536k Description The generalized Fibonacci sequence is defined as follows: R0=a; R1= b; rn= uRn-1 + vRn-2 (n >= 3) Here a,b,u,v are real numbers, called RN for the generalized Fibonacci sequence. Now let you calculate the number of bits of the generalized Fibonacci sequence.

[Leetcode/javascript] 461.Hamming Distance Topic The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Given two integers x and y, calculate the Hamming distance. Note:0≤x, y Example: input:x = 1, y = 4 Output:2 Explanation:1 (0 0 0 1)4 (0 1 0 0)? ? The above arrows point to positions where the corresponding bits are different. Analysis The topic is very simple, given two 32-bit integers

Leetcode 461. Hamming Distance Topics The Hamming distance between and integers is the number of positions at which, the corresponding bits is different. Given integers x and y, calculate the Hamming distance. Solution: Class Solution {public : int hammingdistance (int x, int y) { int z = x^y; int ans = 0; while (z) { ans++; Z = z-1; } return ans; } ; Writing this blog is mainl

1. The array name is a pointer constant representing the first address of the array space, and the constant assignment is not allowed in the program.such as int a[]; A is the array name, which represents the pointer constant of the first address of the array controlA = 0; is wrong and does not allow assigning values to pointer constants2. You can point to a constant by the pointer, but you cannot modify the contents of the constant through the pointersuch as: Char *p;p = "Testing"*p = ' 0 ';3.So

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