先上题目: Continued Fraction Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 332    Accepted Submission(s): 106 Problem Description Dumbear loves numbers very much.One day Dumbear found that each n…

1638: Continued Fraction Time Limit: 1 Sec  Memory Limit: 128 MB Description Input Output Sample Input 4 3 5 1 1 2 5 2 2 Sample Output 11 0 5 30 4 6 1 27 HINT   Source 解题:主要任务是把任一一个分数化成连分式.   方法就是分子分母同时不断的除以分子,直到分子为0.   至于加,减,乘,除都是先算出分数,然后把分数化成连分数..…

题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=6223 题意:给定长度为n的一串数字S,现在要按照一种规则寻找长度为n的数字串,使得该数字串的字典序最大.规则:从数字串S的某一个下标为x的数字出发,可以到达的下一个数字是下标为(x*x+1)%n的数字. 思路:BFS+剪枝.剪枝技巧: 1:遍历到某一层的节点时,记录已经到达过的节点,下次如果还经过就直接不考虑. 2:当前遍历到的某一层节点的数字较之前的小,直接不考虑. AC代码: #define _…

题意:给定一个长度为n(n<=150000)的字符串,每个下标i与(i*i+1)%n连边,求从任意下标出发走n步能走出的字典序最大的字符串. 把下标看成结点,由于每个结点有唯一的后继,因此形成的是外向基环树森林,相当于求基环树上字典序最大的路径. 实际上基环树和树一样是可以通过倍增得到走2^k步能走到的结点的,然后后缀数组又可以通过倍增求出长度为2^k的后缀的字典序,两者结合一下就可以对所有路径按字典序排序了. 复杂度$O(nlogn)$,感觉这应该是标解吧,不过由于数据可能比较随机因此直接BF…

A continued fraction of height n is a fraction of form . You are given two rational numbers, one is represented as  and the other one is represented as a finite fraction of height n. Check if they are equal. Input The first line contains two space-se…

B. Continued Fractions time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A continued fraction of height n is a fraction of form . You are given two rational numbers, one is represented as  a…

任意门:http://codeforces.com/gym/100641/attachments Con + tin/(ued + Frac/tions) Time Limit: 3000/1000 MS (Java/Others)     Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 217     Accepted Submission(s): 27 Description The (simple) contin…

题目链接 现在做这个题目真是千万只草泥马在心中路过 这个与上面一题差不多 这个题目是求e的第100个分数表达式中分子的各位数之和 What is most surprising is that the important mathematical constant,e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...]. The first ten terms in the sequence of convergents for e are: 2, 3,…

这是一道关于RSA的解密题:首先,我们要明白,通常是公钥加密.私钥解密,私钥签名.公钥验证.这个题目中给出的是一个公钥和一段密文. 刚开始一直以为和验证签名有关,费劲脑汁也想不出来怎么办.下面介绍些思路. 首先,利用openssl分析公钥的格式,获得modulus和expoent. 方法一:利用openssl asn1parse来分析公钥格式 root@bt:~/Desktop# openssl asn1parse -in publickey.pub :d= hl= l= cons: SEQUE…

It is possible to show that the square root of two can be expressed as an infinite continued fraction.  2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213... By expanding this for the first four iterations, we get: 1 + 1/2 = 3/2 = 1.51 + 1/(2 + 1/2) = 7…