HDU 4712Hamming Distance（随机函数运用）

Hamming Distance

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1043    Accepted Submission(s): 394

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

 

      题目大意：给你很多串，最多10^5个串，每个串长度是5，16进制转化为2进制，问你任意两个串抑或得到1的最小的个数。

解题思路： 做题的时候只顾着如何降低复杂度，简单的O(n^2)肯定会超时，最后才得知是随机函数写。结果只可能是0~20.自己写的时候10W次还是WA了，人品不行啊，果断随机100W次A了。 不过听说可以更暴力地直接枚举最前面的10000个也可以A。网络赛要敢于尝试。

题目地址：Hamming Distance

AC代码：

#include<iostream>

#include<cstring>

#include<string>

#include<cmath>

#include<cstdio>

#include<algorithm>

using namespace std;

int a[100005];



int main()

{

    int tes,i,j,k,res,ans;

    scanf("%d",&tes);

    while(tes--)

    {

        int n;

        scanf("%d",&n);

        for(i=0;i<n;i++)

            scanf("%X",&a[i]);  //16进制读取

            

        res=20;  //结果初始为最大20

        for(i=1;i<=1000000;i++)

        {

            j=rand()%n;  //随机函数

            k=rand()%n;

            if(j==k)

                continue;

            ans=0;

            int tmp=a[j]^a[k];  //抑或

            while(tmp)  //抑或算1的个数，保存到ans中

            {

                if(tmp&1)

                    ans++;

                tmp>>=1;

            }

            if(ans<res)

                res=ans;

        }

        cout<<res<<endl;

    }

    return 0;

}



//2453MS 676K