hdu 4712Hamming Distance

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

12345

54321

12345

6789A

BCDEF

0137F

Sample Output

Source

2013 ACM/ICPC Asia Regional Online —— Warmup

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <ctime>
 6 #define N 100010
 7 #define inf 0x3f3f3f3f
 8 using namespace std;
 9 int num[20][20];
10 char ss[N][6];
11 int n;
12 
13 int cal(int x,int y)
14 {
15     int ans=0;
16     for(int i=0;i<5;i++)
17     {
18         int dd1,dd2;
19         if(ss[x][i]>='0' && ss[x][i]<='9')
20         {
21             dd1=ss[x][i]-'0';
22         }
23         else
24         {
25             dd1=ss[x][i]-'A'+10;
26         }
27         if(ss[y][i]>='0' && ss[y][i]<='9')
28         {
29             dd2=ss[y][i]-'0';
30         }
31         else
32         {
33             dd2=ss[y][i]-'A'+10;
34         }
35         ans+=num[dd1][dd2];
36     }
37     return ans;
38 }
39 void solve()
40 {
41     srand((unsigned)time(NULL));
42     int res=inf;
43     for(int i=0;i<1000000;i++)
44     {
45         int x=rand()%n;
46         int y=rand()%n;
47         if(x==y)
48         continue;
49         int ans=cal(x,y);
50         res=min(res,ans);
51     }
52     printf("%d
",res);
53 }
54 int main()
55 {
56     for(int i=0;i<16;i++)
57     {
58         for(int j=0;j<16;j++)
59         {
60             int dd=i^j;
61             int len=0;
62             int cnt=0;
63             while(dd>=(1<<len))
64             {
65                 if(dd&(1<<len))
66                 {
67                     cnt++;
68                 }
69                 len++;
70             }
71             num[i][j]=cnt;
72         }
73     }
74     int t;
75     scanf("%d",&t);
76     while(t--)
77     {
78         scanf("%d",&n);
79         for(int i=0;i<n;i++)
80         {
81             scanf("%s",ss[i]);
82         }
83         solve();
84     }
85     return 0;
86 }

View Code

居然是用随机函数就给过了，真厉害，算是学到了一个好方法啊，学习学习