hdu 4712Hamming Distance （厕所大号出的题解）随机函数~

Hamming Distance

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1807    Accepted Submission(s): 715

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

比赛的时候本来想了很久都没思路，   结果和另外一个队友去厕所大号，  聊着聊着我就说，  其实这道题W个400次， 肯定可以过的。  因为才20组，  每组可能20个答案..

本来说笑而已，  结果我们突然发现， 只用生成一个随机函数， 随机选取两个数组的下标， 生成答案，  这样只要随机的次数够多就可以了。  

结果抱着一丝的希望提交，  没想到一交就A了。

#include <cstdio>#include <algorithm>#include <string>#include <cmath>#include <set>#include <map>#include <vector>#include <cstdlib>#include <queue>#include <iostream>#include <functional>#include <cstring>#include <ctime>#define UFOR(i, a, b) for(int i = a; i <= b; i++)#define DFOR(i, a, b) for(int i = a; i >= b; i--)#define MEM0(a) memset(a, 0, sizeof(a))#define MEM1(a) memset(a, -1, sizeof(a))#define MEMINF(a) memset(a, 0x3f3f3f3f, sizeof(a))#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1#define FIN freopen("in.txt", "r", stdin)#define FOUT freopen("out.txt", "w", stdout)using namespace std;//typedef __int64 LL;typedef long long LL;const int MXN = 1e5 + 10;const int MXM = 2e4 + 130;const int HS = 1000007;const int INF = 0x3f3f3f3f;const int MOD = 1000000007;int T, n;int A[MXN];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 < 1e6; 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;}