我的做法,多次宽搜,因为后面的搜索扩展的节点会比较少,所以复杂度还是不需要太悲观的,然后加上一开始对答案的估计,用估计值来剪枝,就可以ac了. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; char a[11]; int key[111111]; int hash[1111111]; int que[2111…

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): 797    Accepted Submission(s): 284 Problem Description (From wikipedia) For binary…

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 Hammi…

Hamming Distance Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 499    Accepted Submission(s): 163 Problem Description (From wikipedia) For binary strings a and b the Hamming distance is equal…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4712 题目大意:任意两个数按位异或后二进制中含1的个数被称为海明距离,给定n个数,求出任意其中两个最小的海明数,输入是16进制. Sample Input 2 2 12345 54321 4 12345 6789A BCDEF 0137F   Sample Output 6 7 分析:用随机数来模拟,加srand()函数,放置产生伪随机数.代码有2个. 代码1: # include<iostream…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4712 解题报告: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…

d.汉明距离是使用在数据传输差错控制编码里面的,汉明距离是一个概念,它表示两个(相同长度)字对应位不同的数量, 我们以d(x,y)表示两个字x,y之间的汉明距离.对两个字符串进行异或运算,并统计结果为1的个数,那么这个数就是汉明距离. 给出N个串,求出其中最小的汉明距离(其中某2个串的汉明距离是最小的). s.随机数法... c.能不能过,看脸... #include<iostream> #include<stdio.h> #include<string.h> #inc…

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 bina…

Hamming Distance Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 1127 Accepted Submission(s): 425 Problem Description (From wikipedia) For binary strings a and b the Hamming distance is equal to t…

Hamming Distance Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 1569    Accepted Submission(s): 616 Problem Description (From wikipedia) For binary strings a and b the Hamming distance is equal…