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[Mathimatics-Numerical algorithms] Ex24PointsGame DL : 0: 一、前言 24点游戏是一个常见游戏，出题者给出4个整数，要求答题者在指定时间内给出一个四则运算的表达式，恰好用上这这个整数各一次，计算结果为24，超出时间为输。二、分析用计算机来算这个题，搜索速度当然有很大优势，我编程喜欢考虑通用一点，不限制输入数量和结果数，甚至不限制运算符数量。这样组合数就很大，如果输入数比较多，则搜索时间会非常长。我用两个方法来提高搜索速度：一、是大家都能考虑到的重复搜索问题，比如1，2，3和2，3，1所有的组合情况是相同的，我只搜索使用递增序的数组，则可以降低一个组合数的数量级别；二、使用动态规划中的备忘录方法，比如你计算出2和3所有可能的计算结果，则他们与4结合的时候，要用到，与1结合的时候，也要用到，使用备忘录，可以只计算一次，大大降低运算复杂度。三、设计整体设计：分别设计4个类：游戏、表达式、运算、分数，各司其责，结构清晰，易于扩展。 -First, the preamble 24-point game is a common game, give the title are four integer requiring answer within a specified period of time in a four calculations give the expression, just use that the integral of the time, results for the 24, beyond the time to lose. Second, analysis Use computers to count the title, the search speed of course, there is a great advantage, I enjoy taking generic programming that does not restrict the importation of the volume and the number of results, and even limiting the number of operators. So on a large number of combinations, if you type a few more, then the search will be a very long time. I used two methods to improve search speed: one is the U.S. able to take into account the problem of duplicate search, such as 1,2,3 and 2,3,1 is a combination of all the same, I just search using the sequence of incremental array, you can reduce the quantity of a combination of the number of level two, the use of dynamic programming methods in the mem
Date : 2026-01-07 Size : 135kb User : xycheng
[Mathimatics-Numerical algorithms] svm4 DL : 0: 　-s svm类型：SVM设置类型(默认0) 　　0 -- C-SVC 　　1 --v-SVC 　　2 – 一类SVM 　　3 -- e -SVR 　　4 -- v-SVR 　　-t 核函数类型：核函数设置类型(默认2) 　　0 – 线性：u v 　　1 – 多项式：(r*u v + coef0)^degree 　　2 – RBF函数：exp(-r|u-v|^2) 　　3 –sigmoid：tanh(r*u v + coef0) 　　-d degree：核函数中的degree设置(针对多项式核函数)(默认3) 　　-g r(gama)：核函数中的gamma函数设置(针对多项式/rbf/sigmoid核函数)(默认1/ k) 　　-r coef0：核函数中的coef0设置(针对多项式/sigmoid核函数)((默认0) 　　-c cost：设置C-SVC，e -SVR和v-SVR的参数(损失函数)(默认1) 　　-n nu：设置v-SVC，一类SVM和v- SVR的参数(默认0.5) 　　-p p：设置e -SVR 中损失函数p的值(默认0.1) 　　-m cachesize：设置cache内存大小，以MB为单位(默认40) 　　-e eps：设置允许的终止判据(默认0.001) 　　-h shrinking：是否使用启发式，0或1(默认1) 　　-wi weight：设置第几类的参数C为weight*C(C-SVC中的C)(默认1) 　　-v n: n-fold交互检验模式，n为fold的个数，必须大于等于2--s svm_type : set type of SVM (default 0) 0-- C-SVC 1-- nu-SVC 2-- one-class SVM 3-- epsilon-SVR 4-- nu-SVR -t kernel_type : set type of kernel function (default 2) 0-- linear: u *v 1-- polynomial: (gamma*u *v+ coef0)^degree 2-- radial basis function: exp(-gamma*|u-v|^2) 3-- sigmoid: tanh(gamma*u *v+ coef0) 4-- precomputed kernel (kernel values in training_instance_matrix) -d degree : set degree in kernel function (default 3) -g gamma : set gamma in kernel function (default 1/k) -r coef0 : set coef0 in kernel function (default 0) -c cost : set the parameter C of C-SVC, epsilon-SVR, and nu-SVR (default 1) -n nu : set the parameter nu of nu-SVC, one-class SVM, and nu-SVR (default 0.5) -p epsilon : set the epsilon in loss function of epsilon-SVR (default 0.1) -m cachesize : set cache memory size in MB (default 100) -e epsilon : set tolerance of termination criterion (default 0.001) -h shrinking: whether to use the shrinking heuristics, 0 or 1 (default 1) -b
Date : 2026-01-07 Size : 17kb User : little863
[Mathimatics-Numerical algorithms] zhizheshengcun DL : 0: 问题的提出：17世纪法国数学家加斯帕的一本《数学的游戏问题》描述了许多有趣问题，约瑟问题就是其中之一： 15名基督教徒和15名异教徒同乘一船航行，途中风浪大作，危机万分，领航者告诉大家，只要将全船的一半人投入海中，其余人就能幸免。大家都同意这个办法，并协定者30人围成一圈；由第一个人起报数，每数至第9人便把他投入海中，下一个接着从1开始报数，第9人又被投入海中，依次循环，直至剩下15人为止。问题是如何排法使投入海中的人全为异教徒？编程的思路：此问题有人把它的答案隐示于下列詩句中： From numbers’ aid and art, (依靠数字的和魔力，) Never will fame depart! （声誉决不会离你而去！）这句诗中元音字母依次为：oueaiaaeeiaeea，分别用1、2、3、4、5代替a、e、i、o、u，便得一排数（划线得数字表示基督教徒，不划线的表示异教徒）。 4 5 2 1 3 1 1 2 2 3 1 2 2 1 故所求排法诗4个基督教徒，5个异教徒，再2个基督教徒，1个异教徒，…可以证明，这种排法所有的基督教徒都是安全的。 -Of the problem: the 17th century French mathematician Gaspard s a "mathematical game problem " describes the many interesting questions, Joseph is one of them: 15 Christians and 15 pagans sail with the boat ride, the way storm comes up, extreme crisis, the leader tell you that as long as half of the whole ship into the sea, the rest can be spared. We all agree with this approach, and there were 30 agreements circle the first person from the reported number, the number to 9 for each person put into the sea in his next start and then reported from a number of people has been put into 9 The sea, in cycles, until the remaining 15 people so far. The question is how to arrange method to make the people into the sea are all infidels? Programming idea: the problem was the answer to it implicitly in the following verse: From numbers aid and art, (relying on numbers and magic) Never will fame depart! (reputation will never leave you!) Poem vowel sentence were: oueaiaaeeiaeea, were re
Date : 2026-01-07 Size : 12kb User : yinlingzhi