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3_6-Ackerman
Ackeman函数的递归方法，运用简洁的递归思想实验复杂数学表达式的计算(Ackeman function recursive method, the use of simple recursive thought experiment to calculate complex mathematical expressions)
- 2015-03-30 16:32:20下载
- 积分：1
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ImageSolarize
image solarizing processing(image processing solarizing)
- 2007-07-10 22:32:24下载
- 积分：1
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ising1.m
1D Ising
ising1(n,m,b,p)
n - length of domain, a positive integer
m - number of iterations, a positive integer
b - coupling parameter / temperature, a positive scalar
b = 0 -> Very high temp / low coupling
b > 1 -> Very low temp / high coupling( 1D Ising ising1 (n, m, b, p) n- length of domain, a positive integer m- number of iterations, a positive integer b- coupling parameter/temperature, a positive scalar b = 0-> Very high temp/low coupling b> 1-> Very low temp/high coupling)
- 2014-10-21 15:31:20下载
- 积分：1
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104
数字移动:图中的九个点上,空出中间的点,其余的点上任意填入数字1到8.1的位置固定不动,然后移动其余的数字,使1到8顺时针从小到大排列.移动的规律是:只能将数字沿线移向空白的点.
(Digital Mobile: Figure in the nine points, empty out the middle point, the rest of the points fill any position of the numbers 1 to 8.1 fixed and then move the rest of the figures, and make 1-8 clockwise from small to large array . The law of movement is: can only move along the number of blank spots.)
- 2009-11-15 21:33:11下载
- 积分：1
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BasicTime
You all might have used CTime or CTimeSpan to manipulate the system timer. Here is an article to show you how your system timer works. I will give you an idea of port communications behind the system clock manipulations. Those who are a little biased on hardware and software interfacing practices would find this article of great help.
- 2007-10-16 09:37:44下载
- 积分：1
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gaussshuzhi
在matlab中用ROMBERG方法求二重积分(Seeking double integral using Romberg method in matlab)
- 2012-04-20 01:03:01下载
- 积分：1
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0
说明： 给定n种物品和一个背包,物品i的重量是Wi,价值是vi,被包容量是C,应该如何选择装入被包的物品,使得装入背包中物品的总价值最大?对于每种物品i,只有两种选择,装入或者不装入.
输入:第一行两个正整数n(物品数)和C(容量)
第二行n个正整数n,表示物品价值
第三行n个正整数n,表示物品重量
输出:第一行表示总价值
第二行n个数(0/1),0表示物品不装入,1表示装入
例如,
输入:
5 10
6 3 5 4 6
2 2 6 5 4
输出
15
1 1 0 0 1
(Species of a given n items and a backpack, the weight of item i is Wi, the value of vi, was packet capacity is C, should be how to choose a package of items to be loaded, the backpack load in the largest total value of goods? For each item i, there are only two options, loaded or not loaded. Input: the first line of the two positive integer n (the number of items) and C (capacity) of the second line of n positive integer n, that the value of the third line of n positive integer n, that the weight of output: the total value of the first line of the second line, said n number of (0/1), 0 that the items do not load, that load 1 for example, enter: 5,106,354,622 6 5 4 output 15 1 1 0 0 1)
- 2009-07-01 23:07:01下载
- 积分：1
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codestatics_shujujiegou
统计代码行、注释行、空行、同时统计函数个数和行数，将代码、注释和空行进行风格评分
(Statistical lines of code, comment lines, blank lines, while the number of statistical functions and the number of rows, the code, comments and blank lines carry style ratings)
- 2013-12-14 09:50:56下载
- 积分：1
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DOA-estimation-
宽带相干信源DOA估计及其阵列校正方法 (Broadband DOA estimation of coherent sources and array calibration method)
- 2013-12-12 13:24:25下载
- 积分：1
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Main_Frequency_decoup_Opti_cdm1
悬置功能计算的，能量解耦的方法，优化及其位移的计算(The calculation of suspended function, energy decoupling method, optimization and calculation of displacement)
- 2015-08-06 09:25:49下载
- 积分：1
