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ConsoleApplication1

于 2015-12-03 发布文件大小:28KB

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下载积分: 1 下载次数: 28

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代码说明：

C#写的两个简单的小程序，分别用牛顿插值法和三次样条插值法来进行插值。(A program written by C#. It uses Newton Interpolation Method and Spline Interpolation Method )

文件列表：

ConsoleApplication1
...................\ConsoleApplication1
...................\...................\bin
...................\...................\...\Debug
...................\...................\...\.....\ConsoleApplication1.exe,7168,2015-12-03
...................\...................\...\.....\ConsoleApplication1.pdb,17920,2015-12-03
...................\...................\...\.....\ConsoleApplication1.vshost.exe,11600,2015-12-03
...................\...................\...\.....\ConsoleApplication1.vshost.exe.manifest,490,2010-03-17
...................\...................\...\Release
...................\...................\ConsoleApplication1.csproj,2459,2015-12-03
...................\...................\ConsoleApplication1.csproj.user,143,2015-12-03
...................\...................\obj
...................\...................\...\x86
...................\...................\...\...\Debug
...................\...................\...\...\.....\ConsoleApplication1.csproj.FileListAbsolute.txt,524,2015-12-03
...................\...................\...\...\.....\ConsoleApplication1.exe,7168,2015-12-03
...................\...................\...\...\.....\ConsoleApplication1.pdb,17920,2015-12-03
...................\...................\...\...\.....\DesignTimeResolveAssemblyReferencesInput.cache,5792,2015-12-03
...................\...................\...\...\.....\TempPE
...................\...................\...\...\Release
...................\...................\Program.cs,5201,2015-12-03
...................\...................\Properties
...................\...................\..........\AssemblyInfo.cs,1370,2015-12-03
...................\ConsoleApplication1.sln,899,2015-12-03

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2011-05-23 11:52:50下载

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2017-04-25 16:06:25下载

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实现高斯厄米特积分，可以直接更换f函数，得到所求的积分(High the Sie Mitt integral to replace the f function, ask the integral )

2021-03-06 20:49:30下载

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2019-03-21 09:39:57下载

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说明： Description 设a[0…n-1]是一个包含n个数的数组，若在ia[j]，则称(i, j)为a数组的一个逆序对（inversion）。比如有5个逆序对。请采用类似“合并排序算法”的分治思路以O(nlogn)的效率来实现逆序对的统计。一个n个元素序列的逆序对个数由三部分构成：（1）它的左半部分逆序对的个数，（2）加上右半部分逆序对的个数，（3）再加上左半部分元素大于右半部分元素的数量。其中前两部分（1）和（2）由递归来实现。要保证算法最后效率O(nlogn)，第三部分（3）应该如何实现？此题请勿采用O(n^2)的简单枚举算法来实现。并思考如下问题：（1）怎样的数组含有最多的逆序对？最多的又是多少个呢？（2）插入排序的运行时间和数组中逆序对的个数有关系吗？什么关系？输入格式第一行：n，表示接下来要输入n个元素，n不超过10000。第二行：n个元素序列。输出格式逆序对的个数。输入样例 5 2 3 8 6 1 输出样例 5(Set a[0... N-1] is a n array containing n numbers. If there is a [i] > a [j] i n the case of I < j, then (i, j) is a n inversion pair of a array. For example, has five reverse pairs. Please use the idea of "merge sorting algorithm" to achieve the statistics of inverse pairs with O (nlogn) efficiency. The number of inverse pairs of a sequence of n elements consists of three parts: (1) The number of reverse pairs in the left half, (2) the number of reverse pairs in the right half, (3) the number of elements in the left half is greater than that in the right half. The first two parts (1) and (2) are implemented by recursion. To ensure the final efficiency of the algorithm O (nlogn), how should the third part (3) be implemented? Do not use O (n ^ 2) simple enumeration algorithm to solve this problem.)

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