http://acm.hdu.edu.cn/showproblem.php?pid=4712

Hamming Distance

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 1610    Accepted Submission(s): 630

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

2
2
12345
54321
4
12345
6789A
BCDEF
0137F

Sample Output

6
7

Source

2013 ACM/ICPC Asia Regional Online —— Warmup

分析：

输入n个数，用十六进制的方式输入的，任意选择其中的两个数进行异或，求异或后的数用二进制表示后1的个数最小的是多少？（n<=100000)

这题看了解题报告，大家都说用随机算法，试过了，随机100000次就过了，50000次都不行，但还是不懂这样怎么可以，唯一的解释就是这个值域也就是结果一共只有21个，

得出正确的结果的可能性很大，但是并不能100%保证结果是对的。无语第一次碰见这种算法。

首先，算汉明距离就是二进制异或以后的1的个数，统计1的个数用x&=x-1很快很神奇。

用if(x&1)  {count++;    x>>=1;}  在位数比较多的时候会慢一些。

然后就是看题解学到的神奇的“随机”！  来取到“任意的两个”  1w次wa，但是10w次就不会，20组testcase ，不会超时。

队友用随机函数在hduoj上交了五次（WA了4次）A了。也是醉啦 ，，，

AC代码：

 1 #include<iostream>
 2 #include<cstring>
 3 #include<string>
 4 #include<cmath>
 5 #include<cstdio>
 6 #include<algorithm>
 7 using namespace std;
 8 int a[100005];
 9
10 int main()
11 {
12     int tes,i,j,k,res,ans;
13     scanf("%d",&tes);
14     while(tes--)
15     {
16         int n;
17         scanf("%d",&n);
18         for(i=0;i<n;i++)
19             scanf("%X",&a[i]);  //16进制读取
20
21         res=20;  //结果初始为最大20
22         for(i=1;i<=1000000;i++)
23         {
24             j=rand()%n;  //随机函数
25             k=rand()%n;
26             if(j==k)
27                 continue;
28             ans=0;
29             int tmp=a[j]^a[k];  //抑或
30             while(tmp)  //抑或算1的个数，保存到ans中
31             {
32                 if(tmp&1)
33                     ans++;
34                 tmp>>=1;
35             }
36             if(ans<res)
37                 res=ans;
38         }
39         cout<<res<<endl;
40     }
41     return 0;
42 }