HDU 472 Hamming Distance （随机数）

Hamming Distance

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1127 Accepted Submission(s): 425

Problem Description

(From wikipedia) For binary strings a and b the Hamming distance is equal to the number of ones in a XOR b. For calculating Hamming distance between two strings a and b, they must have equal length.
Now given N different binary strings, please calculate the minimum Hamming distance between every pair of strings.

Input

The first line of the input is an integer T, the number of test cases.(0<T<=20) Then T test case followed. The first line of each test case is an integer N (2<=N<=100000), the number of different binary strings. Then N lines followed, each of the next N line is a string consist of five characters. Each character is '0'-'9' or 'A'-'F', it represents the hexadecimal code of the binary string. For example, the hexadecimal code "12345" represents binary string "00010010001101000101".

Output

For each test case, output the minimum Hamming distance between every pair of strings.

Sample Input

2212345543214123456789ABCDEF0137F

Sample Output

67

 

题意：就是任意两个16进制字符串XOR，在 所有结果中，求结果以二进制表示时，含1个数最少是多少。

import java.io.*;import java.util.*;public class Main {BufferedReader bu;PrintWriter pw;int t,n,min;public static void main(String[] args) throws IOException{new Main().work();}void work() throws IOException{bu=new BufferedReader(new InputStreamReader(System.in));pw=new PrintWriter(new OutputStreamWriter(System.out),true);t=Integer.parseInt(bu.readLine());while(t--!=0){n=Integer.parseInt(bu.readLine());String str[]=new String[n];for(int i=0;i<n;i++){str[i]=bu.readLine();}min=0xfffffff;Random da=new Random(1);for(int i=1;i<1000000;i++){int x=da.nextInt()%n;int y=da.nextInt()%n;if(x!=y&&x>=0&&y>=0){int num1=Integer.parseInt(str[x],16);int num2=Integer.parseInt(str[y],16);int num=num1^num2;getCount(num);}}pw.println(min);}}void getCount(int num){int len=0;int count=0;while((num>=(1<<len))){if((num&(1<<len))!=0){count++;}len++;}min=Math.min(min,count);}}