从a变到b,也就是将a一直除素因子,除到1为止,然后乘b的素因子,一直乘到b. 但是gcd(a,b)部分是不用除下去的.所以d(a,b)=a/gcd(a,b)的素因子个数+b/gcd(a,b)的素因子个数. 然后....脑洞开始...... 枚举这个因子P,然后去计算a/P的素因子个数+b/P的素因子个数 最小值,也就是要在集合中寻找到一个b,使得式子取得最小值. 然后....可以搞一个数组 c[因子][素因子个数]=多少种情况 来存储集合中的信息. 然后删除,插入的时候维护这个数组即可.具体看…

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 equa…

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…

题意:输入一棵树,输出前k小的点对最短距离dis(i,j)的和. 模拟,官方题解说得很清楚了.不重复了. http://bestcoder.hdu.edu.cn/ 需要注意的是,复杂度要O(n+k),不能用set,map之类的标记是否访问. 一开始TLE了,去掉标记后wa了.最后发现对队列的元素加个前缀,就可以了,即标记该条边是从哪个点延伸的. #include <cstdio> #include <cstring> #include <iostream> #inclu…

Hamming Distance Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 1845    Accepted Submission(s): 740 Problem Description (From wikipedia) For binary strings a and b the Hamming distance is equa…

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…

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…

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…

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…

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…