搜索资源列表
scan
- 本实例展示了并行前缀求和（也称作“scan”）的一种高效的 CUDA 实施途径。针对一个给定的数列，scan 可以计算出一个新的数列，其中每一个元素均为在输入列中该元素前面的所有元素的总和。 -This example shows the parallel prefix sum (also known as " scan" ) of an efficient CUDA implementation of the approach. For a given series, scan
two_int_plus
- C/C++字符串模拟两个大正整数相加，可以改进。-C/C++ string to simulate two positive integers the sum ，code can be improved.
c++
- 用c++代码实现两个至少千位数的大数相加 并在控制台显示输出结果-With c++ Code to achieve at least a thousand two-digit sum of large numbers and in the console display output
polyadd
- 数据结构用数组实现了两个多项式相加，用数组实现多项式的存储-Array data structure implementation using the sum of two polynomials, the polynomial storage array implementation
Problem001
- Find the sum of all the multiples of 3 or 5 below 1000.
PolynomialAdd
- 多项式相加的模拟程序 在控制台环境下运行 适合于初学者-Polynomial sum of the simulation program running in the console environment, suitable for beginners
discrete_convolution_sum
- Implement discrete convolution sum in C
NumericalSummation
- Produce a table of the values of the series (1) for the 3001 values of x, x = 0.0, 0.1, 0.2, …, 300.00. All entries of the table must have an absolute error less than 10-10. This problem is based on a problem from Hamming (1962), when main
godbaha
- 哥德巴赫猜想，输入一个正整数，把它写成两个素数相加的形式-Goldbach conjecture, enter a positive integer, it is written as a sum of two primes
ProjectPT
- The program reads n-size matrix from text file, Finds the upper diagonal matrix. Finds the sum of the elements for each column and finds the maximum of each column. In this project explained opening and cloding text files.-The program reads n-size ma
shurufanxiangxiangjia
- 输入数反相加的C语言程序，对初学者有用。例如输入12+123结果为 21+321 =342-Enter the sum of the number of anti-C language program, useful for beginners. For example, enter 12+123 results 21+321 = 342
Post-Office
- 算法设计与分析中邮局选址问题。 select(int[] array, int begin, int end, int middle)将n个输入元素划分成n/5个组，每组5个 元素，只可能有一个组不是5个元素。用任意一种排序算法，将每组中的元素排好序，并取出每组的中 位数，共n/5个。找出这n/5个元素的中位数。如果n/5是偶数，就找它的2个中位数中较大的一个。 以这个元素作为划分基准。在调用时，只需要将middle参数传入array.length/2就能找到整个数组中的
TaskDistribution
- 算法设计与分析中的任务分配问题。 用回溯法解该问题时，用完全n叉树表示解空间。可行性条件select 剪去不满足行、列约束的子树。递归方法 backtrack(1)实现对整个解空间的回溯搜索；backtrack(i)搜索解空间中的第i层子树，sum记录当前已找到的可行方案 数。在算法backtrack中，当i>n时，算法搜索至叶子结点，当前已找到的可行方案数sum加1；当i<=n时，当前扩展结点 Z是解空间中的内部结点。该结点有x[i]=1,2,...,n共n个
ex1
- This is the code which calculates the sum of the reciprocal values of the numbers in the range of 1 to 20th Do not enter anything over the keyboard. The calculated sum is displayed. The code is written in C + + programming language.
ex2
- This program is in C + + and it calculates the sum of the geometric progression of n members: 1/3 + 1/9 + 1/27 + ... The number of members of the geometric progression n is entered via the keyboard, while the program should be sufficient number of
ex3
- C + + program that will examine all perfect numbers in the range from 1 to 5000. The output of the screen shall be in the form: "Perfect numbers in the range from 1 to 5000 are as follows: ...." Instruction: The number is perfect if it is equal to
ex5
- C + + program to calculate the N-th member of the Fibonacci sequence and present the N-th member on the screen with the Fibonacci range. The number of members of the Fibonacci sequence N, is entered through the keyboard. Fibonacci retracement sequ
Huffman_code
- Huffman编码 1、将信号源的符号按照出现概率递减的顺序排列。 2、将两个最小出现概率进行合并相加，得到的结果作为新符号的出现概率。 3、重复进行步骤1和2直到概率相加的结果等于1为止。 4、在合并运算时，概率大的符号用编码0表示，概率小的符号用编码1表示。 5、记录下概率为1处到当前信号源符号之间的0，l序列，从而得到每个符号的编码。-Huffman coding 1. Source symbols are arranged in order of decreasing
c++LIANXI
- 求两个数的和,C++程序实现，刚刚开始学习，变了简单程序练习。(calculate the sum of two numbers)