Description: 大整数乘法的分治法源码，在JDK1.51中运行-large integer multiplication of the divide-and-conquer method source code, to run in JDK1.51 Platform: |
Size: 1157 |
Author:展丹 |
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Description: LC-3 assembly language machine code to test and solve the Problem of Integer Divide and Modulo. Platform: |
Size: 2048 |
Author:eden |
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Description: 顺序表的结构没有给出，作为演示分治法这里从简顺序表取一整形数组数组大小由用户定义，数据随机生成。我们知道如果数组大小为 1 则可以直接给出结果，如果大小为 2则一次比较即可得出结果，于是我们找到求解该问题的子问题即: 数组大小 <= 2。到此我们就可以进行分治运算了，只要求解的问题数组长度比 2 大就继续分治，否则求解子问题的解并更新全局解以下是代码-e order of the table structure did not give a simple order form as a presentation divide and conquer method to take an integer array array size defined by the user, data is randomly generated. We know that if the array size is 1, you can directly give the results of size of 2, a comparative to the outcome, so we find the sub-problem of solving this problem: the array size <= 2. This is where we can divide and conquer operation, as long as the length of the array of problem solving than the two large, continue to divide and conquer, or to solve the subproblem solution and update the global solution of the following is the code
Platform: |
Size: 1024 |
Author:lzy |
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Description: 算法分析与设计
用递归分治算法解决大整数乘积问题（java源码）-Algorithm Analysis and Design
Recursive divide and conquer algorithm to solve the problem of large integer
multiplication.
(Java source code) Platform: |
Size: 1024 |
Author:jing |
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Description: 这个是一个C++ 代码，功能是输入四个正整数，根据这四个正整数的任意排列，每个整数只能够使用一遍，任意使用加减乘除，找到一个可以计算成24的方法，输出计算方法(This is a C++ code function is to input four positive integers, according to any of the four positive integer permutation, each integer can be used only once, any use to find a add, subtract, multiply and divide, calculated as 24 method, calculation method canbe output) Platform: |
Size: 7168 |
Author:张亚成张亚成
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