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Description: 一个N位数的各位数的N次方之和等于这个数，称为armstrong数（如：153=1^3+5^3+3^3)。编一个程序，求出2，3，4位数中所有的armstrong数。-an N to the median number of Nth power equal to the sum of the number called armstrong (such as : 153 = a ^ ^ 3 5 3 3 ^ 3). Procedures for a series to get two, three or four all of the median armstrong few.
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Size: 1317 |
Author: summer |
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Description: 一个N位数的各位数的N次方之和等于这个数，称为armstrong数（如：153=1^3+5^3+3^3)。编一个程序，求出2，3，4位数中所有的armstrong数。-an N to the median number of Nth power equal to the sum of the number called armstrong (such as : 153 = a ^ ^ 3 5 3 3 ^ 3). Procedures for a series to get two, three or four all of the median armstrong few.
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Size: 1024 |
Author: summer |
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Description: 函数求和：不必给出带参数的具体个数，能求系列数据的和。-Function sum: do not give the specific number of band parameters can order data series and.
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Size: 14336 |
Author: |
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Description: 本人网上搜集的C++笔试题系列，一共十四篇。有微软、华为、中兴等公司的经典笔试题，好东西大家共享。-I-line collection of C++ sum test series, a total of 14. Has Microsoft, Huawei, ZTE and other company' s classic pen test, good things for all to share.
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Size: 202752 |
Author: kevin |
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Description:
在本文件中，哥德巴赫猜想（ 1 ， 1 ） ，证明了复变
一体化。为了证明这猜想，一个新的功能引入Dirichlet级数。
然后，用门阶公式的Dirichlet级数和留数定理，我们
得出这样的结论：任何较大甚至整数可表示为两个素数之和。-In this paper, the Goldbach Conjecture﹛1, 1﹜ is proved by the complex variable
integration. To prove the conjecture, a new function is introduced into Dirichlet series.
And then, by using the Perron Formula of Dirichlet Series and the Residue Theorem, we
conclude that any larger even integer can be expressed as the sum of two primes.
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Size: 84992 |
Author: lll |
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Description: 将正整数n表示成一系列正整数之和：n=n1+n2+…+nk，其中n1≥n2≥…≥nk≥1，k≥1。
正整数n的这种表示称为正整数n的划分。求正整数n的不同划分个数。
例如正整数6有如下11种不同的划分：
6；
5+1；
4+2，4+1+1；
3+3，3+2+1，3+1+1+1；
2+2+2，2+2+1+1，2+1+1+1+1；
1+1+1+1+1+1。-Positive integer n will be expressed as the sum of a series of positive integers: n = n1+ n2+ ...+ nk, of which n1 ≥ n2 ≥ ... ≥ nk ≥ 1, k ≥ 1. Positive integer n such that n is called the division of positive integer. For positive integer n the number of different division. Positive integer 6, for example, the following 11 kinds of different division: 6 5+1 4+2,4+1+1 3+3,3+2+1,3+1+1+1 2+2+ 2,2+2+1+1,2+1+1+1+1 1+1+1+1+1+1.
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Size: 1024 |
Author: yuecuiping |
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Description: 基于多核的并行算法设计，计算一个连续数列的和-The parallel algorithm based on multi-core design, computing a continuous series and
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Size: 2406400 |
Author: tstao |
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Description: 用深度优先搜索。
深度优先搜索的基本思想是：从初始节点开始，在其子节点中选择一个节点进行考察，若不是目标节点，则再在该子节点的子节点中选择一个节点进行考察，一直如此向下搜索。当到达某个子节点，且该子节点既不是目标节点又不能继续扩展时，才选择其只弟节点进行考察。-A ring is composed of n circles as shown in diagram. Put natural numbers into each circle separately, and the sum of numbers in two adjacent circles should be a prime.
Note: the number of first circle should always be 1.
Input
n (0 < n <= 16)
Output
The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements.
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Size: 23552 |
Author: vivian |
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Description: 该乘法器是由8位加法器构成的以时序方式设计的8位乘法器。
其乘法原理是：乘法通过逐项移位相加原理来实现，从被乘数的最低位开始，若为1，则乘数左移后与上一次的和相加；若为0，左移后以全零相加，直至被乘数的最高位。-The multiplier is 8-bit adder consisting of time-series design to the 8-bit multiplier. The multiplication principle is: the sum of multiplication through the principle of each shift to achieve, from the beginning of the lowest multiplicand, if 1, then left after the multiplier and the sum of the last if for 0, left after zero-sum in full, until the highest bit multiplicand.
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Size: 103424 |
Author: lsp |
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Description: sum of series pipelined for trimedia
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Size: 1024 |
Author: Waqas Ahmad |
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Description: 题目：有一分数序列：2/1，3/2，5/3，8/5，13/8，21/13...求出这个数列的前20项之和-Title: There is a sequence of scores: 2/1, 3/2, 5/3, 8/5, 13/821/13 ... find this series the sum of the first 20
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Size: 2048 |
Author: zhang |
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Description: 8数码问题的Java求解，采用A*算法。数据结构采用的Vector<int[]>。输出为"no solution"或一系列表示求解过程的字符串。
某些30步的测试样例需要半分钟左右，但是绝对是正确解~~万望您能不吝点滴时间耐心等待……orz thnx~~
我用的启发函数计算了现在的网格距离目标网格还有多远。
具体算法是，将每一格中的数字的坐标与其应该在的坐标的差加到一起，得到这个最终值h(x)
也就是Manhattan_Distance
源码使用方法：建立Java Project后，分别建立三个同名的.java文件，将源码分别复制过去然后编译运行即可。-a source code that solves the 8 puzzle problem. it uses A-star algorithm, and a major data structure of Vector<int[]>. The output of the program would be "no solution" or a series of strings representing the moves during the solution.
Well, some samples need to be proceeded for around 30 seconds, but the program per se is indeed correct and will ultimately generate the shortest solution. so if u encounter such occasion plz jst wait for a while tho~~~ thnx~!!
the heuristic function, say h(x), in this source code is the sum of the Mahattan Distance of each block from its destination block.
How to use: build a Java Project, and create three .java files with the same names in the .zip file accordingly, and copy the source from the .zip file and just compile and run~~!
have fun thnx~!
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Size: 669696 |
Author: Siegfried |
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Description: MATLAB微积分问题的解析解、函数的级数展开与级数求和问题求解、数值微分、数值积分问题、曲线积分与曲面积分的计算-MATLAB analytical solution of the issue of calculus, the function of the series expansion with the sum of series problem solving, numerical differentiation, numerical integration problems, curve calculation of integral and surface integral
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Size: 2170880 |
Author: qiqi |
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Description: 试验一 级数求和
求当x分别为0.0,0.1,0.2……300.00时级数 的和。要求所有级数和的绝对误差的值都小于1.0×10-10。这一问题是1962年Hamming提出的，即鉴于当时的计算机运行速度非常慢。
-Order to test a number of sum when x ... ... 0.0,0.1,0.2 respectively, when the series and 300.00. Requires that all series and the absolute error values are less than 1.0 × 10-10. The problem is Hamming made in 1962, that view of the then computer is running very slowly.
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Size: 377856 |
Author: 魏青建 |
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Description: c语言下，数列求和实例。
该程序展示了c语言对数列求和-c language, the series sum instance. The program shows the series Sum c language
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Size: 505856 |
Author: Yan |
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Description: This prgm is for finding Sum of Series in VB.
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Size: 1024 |
Author: Arun Sankar |
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Description: 数列求和(动态规划实现)
给定n个数字和一个范围[x,y]，求从这n个数字中任意取出一些数字，使得它们的和在范围[x,y]中有多少种取法。
输入：
输入第一行为整数case，case<=10 表示有case组测试数据。
对于每一组测试数据，第一行为一个整数n (n<=30)，第二行为n个整数a[i]，第三行为两个整数x和y。其中，a[i]>=0，sum(a[i])<2^31，0<x,y<sum(a[i])< span="">。</x,y<sum(a[i])<>
输出：
对于每组数据输出一行，总的取法数。
样例输入：
2
3
1 2 4
1 7
3
1 2 4
2 5
样例输出：
7
4
-Series summation (dynamic programming) given n numbers and a range [x, y], seeking to remove any numbers from these n numbers such that their sum in the range [x, y] how much emulated . Input: input the first acts of integer CASE case < = 10 means that the case set of test data. For each set of test data, the first is an integer n (n < = 30), the second n integers a [i], the third two integers x and y. Wherein a [i]> = 0, the sum (a [i]) < 2 ^ 31,0 <x,y<sum(a[i])< span=""> . </x,y<sum(a[i])<> Output: For each set of data output line, the total number of emulated. Sample Input: 2 3 1 4 1 7 3 1 2 4 2 5 Sample Output: 74
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Author: 陈曦 |
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Description: This code is based on C language, it is used to find th sum of cos(x) series.
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Size: 13312 |
Author: KARTHIKEYAN |
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Description: Fibonacci数列的特点是第1，2两个数为1，1。从第3个数开始，该数是前两个数之和。求这个数列的前30个元素。-Fibonacci sequence is characterized by two numbers 1,2 1,1. The first three numbers from the beginning, the number and the sum of the first two. 30 elements before seeking this series.
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Size: 1024 |
Author: |
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Description: The Fibonacci sequence is a series where the next term is the sum of pervious two terms. The first two terms of the Fibonacci sequence is 0 followed by 1.
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Size: 1024 |
Author: Dharmesh |
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