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Description: 一个将正整数转变成二进制数的代码，可以编译运行-to a positive integer into binary code, the compiler can run
Platform: |
Size: 22766 |
Author: 周远 |
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Description: Jollen-Kit! 2006 是非常積極進行中的專案，所有紀錄與更新將同步揭示於本網頁。
快速下載
U-Boot 的 binary code：U-Boot binary - 2006/01/03
Linux kernel 2.4.18 for jk2410：uimage.img - 2006/01/06
可開機的 Base root filesystem： urootfs.img -2006/01/07 (without tftp)
可開機的 Base root filesystem： urootfs2.img -2006/03/10 (Busybox 1.10 with tftp)
GNU cross toolchain：arm-9tdmi-linux-gnu.tar.gz (armpath by Joe)
安裝方法:
# cd / (切換到 / 根目錄)
# tar jxf <路徑>/arm-9tdmi-linux-gnu.tar.gz (將檔案解開)
\"Hello, World!\" 在 ARM9 的範例：hello_arm.tar.gz (使用Makefile)
其它工具與原始碼下載
驅動程式模組
-Jollen-Kit! 2006 is very positive for the ad hoc, all records will be updated with synchronization revealed on this website. Fast download U-Boot of binary code : U-Boot binary-2005/10/02 Linux kernel 2.4.18 for jk2410 : uimage.img - 2006-01-06 can boot Base root filesys tem : urootfs.img -2006/01/07 (without tftp) boot B Rnase root filesystem : urootfs2.img -2006/03/10 (Busybox 1.10 with t ftp) GNU cross toolchain : arm-9tdmi - linux-piumarta/squeak/unix/release/Squeak-3.1beta-i686-PC-linux-gnu.tar.gz (armpath by Joe) Installation Method : # cd / (switched to the / root directory) # tar jxf
Platform: |
Size: 13768 |
Author: ffllkk |
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Description: ACM试题An Easy Problem
Description
As we known, data stored in the computers is in binary form. The problem we discuss now is about the positive integers and its binary form. Given a positive integer I, you task is to find out an integer J, which is the minimum integer greater than I, and the number of 1 s in whose binary form is the same as that in the binary form of I. For example, if \"78\" is given, we can write out its binary form, \"1001110\". This binary form has 4 1 s. The minimum integer, which is greater than \"1001110\" and also contains 4 1 s, is \"1010011\", i.e. \"83\", so you should output \"83\".
-An Easy Problem Description As we kno wn. data stored in the computers i n binary form. T he problem we discuss now is about the positive i ntegers and its binary form. Given a positive in teger I, you task is to find out an integer J, which is the minimum integer greater than I, and the number of 1's in whose binary form is the sa me as that in the binary form of I. For example, if "78" is given. we can write out its binary form. "1001110." This binary form has 4 1 s. The minimu m integer, which is greater than "1001110" and also contai ns 4 1's, is "1010011", i.e., "83", so you should output "83."
Platform: |
Size: 9443 |
Author: LY |
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Description: Given an positive integer A (1 <= A <= 109), output the lowest bit of A. For example, given A = 26, we can write A in binary form as 11010, so the lowest bit of A is 10, so the output should be 2. Another example goes like this: given A = 88, we can write A in binary form as 1011000, so the lowest bit of A is 1000, so the output should be 8.
-positive integer A (1
Platform: |
Size: 679 |
Author: shao |
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Description: 一个将正整数转变成二进制数的代码，可以编译运行-to a positive integer into binary code, the compiler can run
Platform: |
Size: 22528 |
Author: 周远 |
Hits:
Description: Jollen-Kit! 2006 是非常積極進行中的專案，所有紀錄與更新將同步揭示於本網頁。
快速下載
U-Boot 的 binary code：U-Boot binary - 2006/01/03
Linux kernel 2.4.18 for jk2410：uimage.img - 2006/01/06
可開機的 Base root filesystem： urootfs.img -2006/01/07 (without tftp)
可開機的 Base root filesystem： urootfs2.img -2006/03/10 (Busybox 1.10 with tftp)
GNU cross toolchain：arm-9tdmi-linux-gnu.tar.gz (armpath by Joe)
安裝方法:
# cd / (切換到 / 根目錄)
# tar jxf <路徑>/arm-9tdmi-linux-gnu.tar.gz (將檔案解開)
"Hello, World!" 在 ARM9 的範例：hello_arm.tar.gz (使用Makefile)
其它工具與原始碼下載
驅動程式模組
-Jollen-Kit! 2006 is very positive for the ad hoc, all records will be updated with synchronization revealed on this website. Fast download U-Boot of binary code : U-Boot binary-2005/10/02 Linux kernel 2.4.18 for jk2410 : uimage.img- 2006-01-06 can boot Base root filesys tem : urootfs.img-2006/01/07 (without tftp) boot B Rnase root filesystem : urootfs2.img-2006/03/10 (Busybox 1.10 with t ftp) GNU cross toolchain : arm-9tdmi- linux-piumarta/squeak/unix/release/Squeak-3.1beta-i686-PC-linux-gnu.tar.gz (armpath by Joe) Installation Method :# cd/(switched to the/root directory)# tar jxf
Platform: |
Size: 13312 |
Author: ffllkk |
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Description: 一个二值化的简单工程，初学者可以参考，真的不错哦-a value of two simple, beginners can refer to, oh really good
Platform: |
Size: 2899968 |
Author: wang |
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Description: ACM试题An Easy Problem
Description
As we known, data stored in the computers is in binary form. The problem we discuss now is about the positive integers and its binary form. Given a positive integer I, you task is to find out an integer J, which is the minimum integer greater than I, and the number of 1 s in whose binary form is the same as that in the binary form of I. For example, if "78" is given, we can write out its binary form, "1001110". This binary form has 4 1 s. The minimum integer, which is greater than "1001110" and also contains 4 1 s, is "1010011", i.e. "83", so you should output "83".
-An Easy Problem Description As we kno wn. data stored in the computers i n binary form. T he problem we discuss now is about the positive i ntegers and its binary form. Given a positive in teger I, you task is to find out an integer J, which is the minimum integer greater than I, and the number of 1's in whose binary form is the sa me as that in the binary form of I. For example, if "78" is given. we can write out its binary form. "1001110." This binary form has 4 1 s. The minimu m integer, which is greater than "1001110" and also contai ns 4 1's, is "1010011", i.e., "83", so you should output "83."
Platform: |
Size: 234496 |
Author: LY |
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Description: Given an positive integer A (1 <= A <= 109), output the lowest bit of A. For example, given A = 26, we can write A in binary form as 11010, so the lowest bit of A is 10, so the output should be 2. Another example goes like this: given A = 88, we can write A in binary form as 1011000, so the lowest bit of A is 1000, so the output should be 8.
-positive integer A (1
Platform: |
Size: 1024 |
Author: shao |
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Description: 以输入的正整数的值作为二叉排序树中的结点的数据场之值，建立一棵二叉排序树-To enter the positive integer value as a binary sort tree nodes in the data field
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Size: 10240 |
Author: cynthiasong |
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Description: 格雷码生成
对于给定的正整数n，格雷码为满足如下条件的一个编码序列：
(1) 序列由2n个编码组成，每个编码都是长度为n的二进制位串。
(2) 序列中无相同的编码。
(3) 序列中位置相邻的两个编码恰有一位不同。
-Gray code generated for a given positive integer n, Gray code to meet the following conditions of a coding sequence: (1) coding sequence by the 2n months, each code is a length of n-bit binary string. (2) without the same sequence number. (3) the location of sequences adjacent to the two just have a different encoding.
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Size: 1024 |
Author: Rei |
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Description: 转换10进制数为二进制数。
％ B = DE2BI(D) 转换正整数向量D成二进制矩阵B。
二进制矩阵B的每一行表示十进制向量D中相应的数。
B = DE2BI(D, N) 转换正整数向量D成二进制矩阵B，-Conversion of 10 hexadecimal for binary number. B = DE2BI (D) positive integer conversion into a binary vector D matrix B. Binary matrix B that each row vector D decimal number corresponding. B = DE2BI (D, N) vector D converter positive integer binary matrix B,
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Size: 1024 |
Author: 王云 |
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Description: 十进制与二进制之间相互转换 正负数 整数与小数-Decimal and binary conversion between the number of positive and negative integers and decimals
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Size: 270336 |
Author: ze |
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Description: 从键盘上输 入一串正整数， 最后输入－1作为输入结束的标志。如输入的序列为：2，5，7，23，48，96，……，-1。请以这些正整数的值作为二叉排序树中的结点的数据场之值，建立一棵二叉排序树。注意：请采用动态存储方法保存这棵二叉排序树，事先并未知道该二叉排序树中的结点的个数。-Input from the keyboard on a string of positive integers, the last input-1 as an input symbol of the end. Such as input sequence as follows: 2,5,7,23,48,96, ... ...,-1. Please refer to these as a positive integer value of two binary sort tree nodes in the data field' s value, set up a binary sort tree. Note: please use the dynamic storage method to preserve tree binary sort tree, the prior did not know that the binary sort tree in the number of nodes.
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Size: 237568 |
Author: syuanying |
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Description: 格雷码问题
1.问题描述
对于给定的正整数n，格雷码为满足如下条件的一个编码序列：
(1) 序列由2n个编码组成，每个编码都是长度为n的二进制位串。
(2) 序列中无相同的编码。
(3) 序列中位置相邻的两个编码恰有一位不同。
例如：n=2时的格雷码为：{00, 01, 11, 10}。
设计求格雷码的递归算法并实现。
-Gray-code problem 1. Problem description for a given positive integer n, Gray code to meet the following conditions of a coding sequence: (1) coding sequence by the 2n months, each code is a length of n-bit binary string. (2) without the same sequence number. (3) the location of sequences adjacent to exactly two has a different encoding. For example: n = 2 when the Gray code is: (00, 01, 11, 10). Designed for Gray code and implementation of the recursive algorithm.
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Size: 1746944 |
Author: 刘珊珊 |
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Description: 一个四位二进制加/减运算器。
要求：当控制端G＝0时做加运算，G＝1时做减运算。用发光二极管表示运算结果的正、负。用数码管显示运算结果：加运算时，相加之和不超过15，减运算时，结果可正可负，但都用原码表示。-Plus a four binary/by calculator. Requirements: When the control terminal G = 0 when computing increases, G = 1 when computing reduced. Computing with light-emitting diodes, said the results of positive and negative. Digital display computing Results: Canadian operations, the sum of not more than 15, by calculation, the result can be negative now, but they said the original code.
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Size: 224256 |
Author: 张三 |
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Description: 一种改进的RFID二进制搜索防碰撞算法，基于二进制树的防碰撞算法研究-An improved anti-collision RFID binary search algorithm, binary tree-based anti-collision algorithm
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Size: 167936 |
Author: 李伟 |
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Description: 一·数据定义及存储格式：（数据段定义不同数据演示）
数据在数据区的存储，DW、DB数据定义，十进制、二进制、B格式、H格式、BCD码、ASCII码，
正数、负数（补码）存储及汉字存储。
二·段地址与物理地址：数据段DS、ES、堆栈段SS、程序段CS
三·将数据区BUF1中前N个数，分别与数M相加，数据仍保存在原处。-Definition and a data storage format: (the definition of data different data presentation above) data stored in the data area, DW, DB data definition, decimal, binary, B format, H format, BCD code, ASCII code positive, negative (complement) storage and storage of Chinese characters. Bis address and physical address: data segment DS, ES, paragraph stack SS, program segment CS three data areas BUF1 in the number of pre-N, respectively, with a number of M add up, data is still stored in the original.
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Size: 2048 |
Author: bob |
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Description: 十进制转二进制，正负数、补码、浮点均可,非常强大-Decimal to binary, positive and negative numbers, complement, floating point can be
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Size: 1024 |
Author: ftj |
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Description: 利用“二值图像与原图像做点乘,得到子图像”的原理.，编写M程序，构造特殊的二值图像，最终得到需要的子图像。
编写M程序，实现图像的平移，要求两个平移量TX、TY可取正负值 。
-By using the principle of multiplying the binary image and the original image to get the sub-image, the M program is programmed to construct a special binary image to obtain the required sub-image.
The preparation of M procedures to achieve the translation of the image, requiring two translations TX, TY desirable positive and negative values.
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Size: 1155072 |
Author: carrie |
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