Welcome![Sign In][Sign Up]
Location:
Search - Prime Ring Problem

Search list

[Embeded-SCM Develops

Description: 6. 素数环问题 把从1到20这20个数摆成一个环,要求相邻的两个数的和是一个素数。 分析:用回溯算法,考察所有可能的排列。 -6. Prime numbers loop problem 1-20 from 20 before the number into a ring, adjacent to both the number of requests and is a prime number. Analysis: The retrospective algorithm, examining all possible permutations.
Platform: | Size: 173056 | Author: 松柏长青 | Hits:

[Data structshangdian1016

Description: Prime Ring Problem 杭州电子科技大学ACM在线判定系统的1016题素数环,本人自己写的算法,有兴趣的同学可以拿去参考下,正常情况是可以AC的-Prime Ring Problem, Hangzhou University of Electronic Science and Technology ACM determine the system' s 1016 title-line prime ring, I write their own algorithms, interested students can take it for reference, the normal situation is AC,
Platform: | Size: 175104 | Author: 无晴 | Hits:

[Data structssuanfa2

Description: 分支限界法、分治法、动态规划法常见代码,素数环问题、背包问题、抄写书稿问题。-Branch and bound method, divide and conquer, dynamic programming method common code, prime ring problem, knapsack problem, the problem copying the manuscript.
Platform: | Size: 41812992 | Author: 李诗诗 | Hits:

[Other1016

Description: Prime Ring Problem, A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime-Prime Ring Problem, A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime
Platform: | Size: 1024 | Author: ouyangyang | Hits:

[OS programNEWJ

Description: 我用C++写的分解质因数算法和排队报数算法。其中排队问题为:若干人站成一个首尾相连的环,从第一个人开始从1到3报数,每次数到3的人站出来,问最后剩下的那个人是原先的第几个人。-I with C++ write the prime factorization algorithm and the number of queued message algorithm. Queuing problem: a number of people standing in a ring of end-to-end, from the first person from 1 to 3 the number of reported each number to 3 people to stand up and ask the last remaining original first few people .
Platform: | Size: 1024 | Author: 汪发进 | Hits:

CodeBus www.codebus.net