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ComputationalMechanicsLibrary
一个个人使用的精简了的c++有限元计算库(Finite element calculation library)
- 2018-04-15 15:39:13下载
- 积分:1
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FENZHI
分支界限费用矩阵的说明文档,非常实用,建议下载哦(Branch line cost matrix documentation, very practical, it is recommended to download Oh)
- 2013-09-13 14:24:50下载
- 积分:1
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newton
数值分析的牛顿法,是用C编写的,希望大家互相参考!!1(Pleiades值tub cavity Xu Yang Lai Hey, Did using C met写cavity, submerged mode希望deceive gaze intently EC banana!! 1)
- 2008-05-14 13:25:28下载
- 积分:1
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floyd
求解最短路算法,,,
赋路径初值 更新 显示迭代步数
显示每步迭代后(solve the lest distance)
- 2013-09-13 12:35:54下载
- 积分:1
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11087 统计逆序对
Description
设a[0…n-1]是一个包含n个数的数组,若在i<j的情况下,有a[i]>a[j],则称(i, j)为a数组的一个逆序对(inversion)。
比如 <2,3,8,6,1> 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, <2,3,8,6,1> 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.)
- 2019-01-07 23:52:06下载
- 积分:1
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jx
说明: 矩形件排样的模拟退火算法求解
适用于矩形件的排样(Rectangular pieces of nesting of simulated annealing algorithm for nesting of rectangular parts)
- 2008-08-20 18:15:39下载
- 积分:1
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damping-ratio-for-silk-and-sand
粉土的阻尼比拟合程序。拟合模型选择Davidenkov模型。(The silt damping match the proper procedure. The Fitting Model Davidenkov model.)
- 2013-05-11 12:01:13下载
- 积分:1
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Map-projection
1、根据所给的中国版图数据
绘制显示基于北京54坐标系的经纬度数据
编写兰勃特投影转换程序,转换上述数据,并显示
编写墨卡托投影转换程序,转换上述数据,并显示
同时,要绘制相对应的经纬网格,网格间距5度
2、根据所给的世界版图数据
绘制显示基于WGS84坐标系的经纬度数据
编写墨卡托投影转换程序,转换上述数据,并显示
计算北京(116.4,39.8)到巴黎(2.2, 48.52)的大圆轨迹,并显示
同时,要绘制相对应的经纬网格,网格间距5度(1, according to the Chinese territory given display data drawn based on latitude and longitude data write Beijing 54 coordinate system Lambert projection conversion process, the conversion of the data, and displays the write Mercator projection conversion process, the conversion of the data, and display the same time, Draw the corresponding latitude and longitude grid, the grid spacing of 5 degrees 2, according to the world map rendering display data written to the Mercator projection based on latitude and longitude data conversion program WGS84 coordinate system, the conversion of the data, and displays the calculation Beijing (116.4, 39.8) to Paris (2.2, 48.52) the great circle track and display the same time, to draw the corresponding latitude and longitude grid, the grid spacing of 5 degrees)
- 2015-12-08 21:51:30下载
- 积分:1
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四阶龙格-库塔法
说明: 利用四阶龙格库塔求解微分方程,并给出方程实例。(The fourth order Runge Kutta is used to solve the differential equation and an example is given.)
- 2020-07-04 18:33:12下载
- 积分:1
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BayesianLinearRegression
This matlab code is a demo version of Bayesian linear regression.
- 2010-02-01 14:09:22下载
- 积分:1