Hamming Distance

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)

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

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其中可能提交会出现错误，但是多提交几次就对了（建议大家该为1e6就可以了）
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <cstdio>
#include <string>
#include <ctime>
using namespace std;
typedef long long LL;
const int MAXN = 1e5 + 5;
const int INF = 0x3f3f3f3f;
int T, n;
int A[MAXN];
int Get_Num(int m) {
    int ret = 0;
    while(m) {
        ret += (m & 1);
        m >>= 1;
    }
    return ret;
}
int main() {
    scanf("%d", &T);
    while(T --) {
        scanf("%d", &n);
        for(int i = 0; i < n; i ++) {
            scanf("%X", &A[i]);
        }
        srand(time(NULL));
        int  Min = INF;
        for(int i = 0; i < 1e5; i ++) {
            int a = rand() % n;
            int b = rand() % n;
            if(a == b) continue;
            Min = min(Min, Get_Num(A[a] ^ A[b]));
        }
        printf("%d\n",Min);
    }
    return 0;
}

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