怎样将内容是binary的字符串转换成byte[]数组

怎样将内容是binary的字符串转换成byte[]数组,并存入sql server image类型字段。

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子字符串,The most binary substring
Problem Description Everybody knows the number is saved with the binary string in the computer. Now, their have N (1 <= N <= 1000) binary strings I tell you, your task is tell me what is the most binary substring with K (1 <= K <= L) characters in the strings I give you. The length of each string is L (1 <= L <= 60). Output times of the most appearing binary substring. Input Each line will contain three numbers N, L, K. Following N lines, represent N binary strings. Output One answer one line. Sample Input 2 6 3 101011 110101 2 6 4 101011 110101 Sample Output 4 2
注册列表的reg_binary二进制的转换规则是?
____我的注册列表的SQ有这些数据 ,fd af da fd cc ce ff af, 1请问转成字符串会是什么内容?_ 关于reg_binary 在字符串和双字不适用的场合,可以使用二进制值。二进制值可以用来储存任意类型的数据,其格式为 hex:xx,yy,zz 等,其中xx、yy、zz是用16进制值表示的单字节值,二进制值可以是任意长度的。 较长的二进制值可以用C语言多行符 / 分割成多行显示。 例如: "bar"=hex:48,00,00,00,01,00,00,00,0a,00,0a,00,0a,00,0a,00,0a,00,0a,00,0a,/ 00,0a,00,0a,00,0a,00,0a,00,0a,00,0a,00,0a,00,0a,00,0a,00,0a,00,0a,00,0a,00,/ 0a,00,0a,00,0a,00,0a,00,0a,00,0a,00,0a,00,0a,00,0a,00,00,00,00,00,c4,ac,01,/ 00 请记住,/ 符只能够应用在二进制值当中。另外,微软声明一个二进制值不可以存储超过2K的资料,这也是值得注意的。 二进制值的例子: "foo"=hex:00,de,ca,de,12,34 2 例子里面是如何算出来的?为什么foo那么短 “bar”那么长?___
Binary Number 二进制的数字
Problem Description For 2 non-negative integers x and y, f(x, y) is defined as the number of different bits in the binary format of x and y. For example, f(2, 3)=1,f(0, 3)=2, f(5, 10)=4. Now given 2 sets of non-negative integers A and B, for each integer b in B, you should find an integer a in A such that f(a, b) is minimized. If there are more than one such integer in set A, choose the smallest one. Input The first line of the input is an integer T (0 < T ≤ 100), indicating the number of test cases. The first line of each test case contains 2 positive integers m and n (0 < m, n ≤ 100), indicating the numbers of integers of the 2 sets A and B, respectively. Then follow (m + n) lines, each of which contains a non-negative integers no larger than 1000000. The first m lines are the integers in set A and the other n lines are the integers in set B. Output For each test case you should output n lines, each of which contains the result for each query in a single line. Sample Input 2 2 5 1 2 1 2 3 4 5 5 2 1000000 9999 1423 3421 0 13245 353 Sample Output 1 2 1 1 1 9999 0
输入字符串构建两个二叉搜索树
A tempting approach to maintaining a balanced binary search tree is to maintain two binary search trees and to insert each new key into the tree that will be more balanced. More specifically, do the following: The first key is the root of the left tree. The second key is the root of the right tree. To add a new key, insert it into the tree where it would have a smaller depth. If the depths of both trees are the same, then add it to the first tree.The input will be 5 strings (ignore everything but the letters A through Z and a through z; uppercase and lowercase are the same). For each input string, build the two trees with each string as described above. Print contents of the first tree in preorder (root, then the left child, then the right child). Sample Input: Line #1: AMERICAN COMPUTER SCIENCE Line #2: I must say that I find television very educational Line #3: INTERMEDIATE DIVISION Sample Output: Output #1: A A E C C E E N I P S Output #2: I A A A A I D D C E E I U T T S R U Y V Y Output #3: I E E A I I I T T V 维护一个平衡的二叉搜索树的一个诱人的方法是维护两个二叉搜索树,并将每个新键插入树中,这样会更加平衡。更具体地说,执行以下操作:第一个键是左树的根。第二个键是右树的根。若要添加新键,请将其插入到树中其深度较小的位置。如果两棵树的深度相同,则将其添加到第一棵树上。 输入将是5个字符串(忽略除了字母A到Z和A到Z;大写和小写是一样的)。对于每个输入字符串,使用上面描述的每个字符串构建两个树。按顺序打印第一个树的内容(根,然后左子树,然后右子树)。 样例输入: 第一行:美国计算机科学 我得说我觉得电视很有教育意义 第3行:中间部分 样例输出: 输出#1:A A E C C E E N I P S 输出#2:I A A A I D D C E E I U T S R U Y V Y 输出#3:ieia I I T T V 基本要求:算法合理,结果准确
C++ 数据转换 如何将文本文件(.txt)转换成二进制文件(.dat)?
有什么方法可以把int 类型的数组转换成 unsigned char的数组? 因为我现在从文本文件读取(0-255)的数据,一行一个数据,读取到在int数组每个数据占用三个字节,我想只占用一个字节,然后保存二进制文件,请问怎么可以解决呢? 例如,文本文件数据是255,这样写入之后二进制文本之后 //ofstream fout("Binary.dat", ios_base::out | ios_base::binary); for (i = 0; i < LINES; i++) //输出二进制文本 { out.write((char*)&BINARY[i], sizeof(BINARY[i])); } 希望读取到的二进制数据还是255。题主在做随机数,需要用Binary,即每BYTE占用八个BITs Lines = CountLines(filename); cout << "文件行数:" << Lines << endl; // 读取文件行数 int *Bit = new int[Lines]; int i = 0; while (!file.eof()) { file >> Bit[i]; i++; } file.close(); //读取文本文件数据到数组,关闭文件 for (i = 0; i < Lines; i++) { cout << Bit[i]<<endl; } delete[]Bit; cin.get(); //输出数组内容到屏幕 目前,我只是把数据读到一个int数组(Bits)里面了。
Binary Number 二进制数字的问题
Problem Description For 2 non-negative integers x and y, f(x, y) is defined as the number of different bits in the binary format of x and y. For example, f(2, 3)=1,f(0, 3)=2, f(5, 10)=4. Now given 2 sets of non-negative integers A and B, for each integer b in B, you should find an integer a in A such that f(a, b) is minimized. If there are more than one such integer in set A, choose the smallest one. Input The first line of the input is an integer T (0 < T ≤ 100), indicating the number of test cases. The first line of each test case contains 2 positive integers m and n (0 < m, n ≤ 100), indicating the numbers of integers of the 2 sets A and B, respectively. Then follow (m + n) lines, each of which contains a non-negative integers no larger than 1000000. The first m lines are the integers in set A and the other n lines are the integers in set B. Output For each test case you should output n lines, each of which contains the result for each query in a single line. Sample Input 2 2 5 1 2 1 2 3 4 5 5 2 1000000 9999 1423 3421 0 13245 353 Sample Output 1 2 1 1 1 9999 0
c++下16进制转到文本字符串
我将图片文件用二进制读取出来存入string中,然后转成了16进制的字符串,方法如下: unsigned char c; char buf[3]; std::string result = ""; std::ifstream fread(file,std::ios::binary); while(fread.read((char*)(&c), sizeof(c))) { sprintf(buf, "%X", c); result += buf; } std::cout << result << std::endl; fread.close(); 我想问下,我应该怎么把它转回string然后写回图片文件啊?
Binary String 的写法
Problem Description Give you a binary string, you can add some 1s or 0s at any places of the string. Then the result divide by K, and the remainder should be zero. Please print string of the result, if there are more, print the one with the smallest length, if still more, print the smallest lexicographical one. If there is no such result print impossible. Input No more than 10 test cases. For each test case first line is a nonempty binary string whose length is no more than 20 and without leading zeros. Next line is an integer K(1 <= K <= 200). Output Print the result described above. Sample Input 1000 4 11101 34 11010 70 Sample Output 1000 10101010 11010010
银联支付接口使用rsa 数据加密。明文,密钥,都是16进制的字符串
银联支付接口使用rsa 数据加密。明文,密钥,都是16进制的字符串,加密结果和对方给的小工具加密的结果不一样 小工具rsatools.exe 求大神指导 代码如下 import java.io.ByteArrayOutputStream; import java.math.BigInteger; import java.security.Key; import java.security.KeyFactory; import java.security.interfaces.RSAPublicKey; import java.security.spec.RSAPublicKeySpec; import javax.crypto.Cipher; import org.apache.commons.codec.binary.Hex; import org.bouncycastle.jce.provider.BouncyCastleProvider; public class test { /** * 加密 * * @param key * 加密的密钥 * @param data * 待加密的明文数据 * @return 加密后的数据 * @throws Exception */ public byte[] encrypt(Key key, byte[] data) throws Exception { try { Cipher cipher = Cipher.getInstance("RSA", new BouncyCastleProvider()); cipher.init(Cipher.ENCRYPT_MODE, key); // 获得加密块大小,如:加密前数据为128个byte,而key_size=1024 加密块大小为127 // byte,加密后为128个byte; // 因此共有2个加密块,第一个127 byte第二个为1个byte int blockSize = cipher.getBlockSize(); int outputSize = cipher.getOutputSize(data.length);// 获得加密块加密后块大小 int leavedSize = data.length % blockSize; int blocksSize = leavedSize != 0 ? data.length / blockSize + 1 : data.length / blockSize; byte[] raw = new byte[outputSize * blocksSize]; int i = 0; while (data.length - i * blockSize > 0) { if (data.length - i * blockSize > blockSize) cipher.doFinal(data, i * blockSize, blockSize, raw, i * outputSize); else cipher.doFinal(data, i * blockSize, data.length - i * blockSize, raw, i * outputSize); // 这里面doUpdate方法不可用,查看源代码后发现每次doUpdate后并没有什么实际动作除了把byte[]放到ByteArrayOutputStream中 // ,而最后doFinal的时候才将所有的byte[]进行加密,可是到了此时加密块大小很可能已经超出了OutputSize所以只好用dofinal方法。 i++; } return raw; } catch (Exception e) { throw new Exception(e.getMessage()); } } /** * 解密 * * @param key * 解密的密钥 * @param raw * 已经加密的数据 * @return 解密后的明文 * @throws Exception */ public byte[] decrypt(Key key, byte[] raw) throws Exception { try { Cipher cipher = Cipher.getInstance("RSA", new org.bouncycastle.jce.provider.BouncyCastleProvider()); cipher.init(cipher.DECRYPT_MODE, key); int blockSize = cipher.getBlockSize(); ByteArrayOutputStream bout = new ByteArrayOutputStream(64); int j = 0; while (raw.length - j * blockSize > 0) { bout.write(cipher.doFinal(raw, j * blockSize, blockSize)); j++; } return bout.toByteArray(); } catch (Exception e) { throw new Exception(e.getMessage()); } } /** * 测试 * * @param args * @throws Exception */ public static void main(String[] args) throws Exception { RSAUtil rsa = new RSAUtil(); String modeHex = "BB83CB428FF41B61762FB8A34041EF8B897301DC4DB84CBD60FEBFC244260B7CC3D57925591970028BB466E13C5A51650213DB7566A78453EA55D725A9B78884A99FD8B1530499D08F3D8CF078CDA4346395CCCA8379E814559E3F6A7DB851C9FCA1FC5A0D3983C637E33B02DA65DF139428C19D9EE2AD8F9A15663E984B166B"; String messageg = "06111111FFFFFFFFA1D93F0C1C682F1836B7E97CC0A8D37363237EA630383632323235383030303030303030303030303120202020202020202036323232353830303030303030303030303032202020202020202020303531323234363536343836352020343231303333202020202020202020202020202020202020202020"; String exponentHex = "010001"; KeyFactory factory = KeyFactory.getInstance("RSA", new BouncyCastleProvider()); BigInteger n = new BigInteger(modeHex, 16); BigInteger e = new BigInteger(exponentHex, 16); RSAPublicKeySpec spec = new RSAPublicKeySpec(n, e); RSAPublicKey pub = (RSAPublicKey) factory.generatePublic(spec); byte[] mi = rsa.encrypt(pub, messageg.getBytes("GB2312")); String result1 = new String(Hex.encodeHex(mi)); System.out.println("加密后==" + result1); } }
Counting Binary Trees 树的算法编程
Problem Description There are 5 distinct binary trees of 3 nodes: Let T(n) be the number of distinct non-empty binary trees of no more than n nodes, your task is to calculate T(n) mod m. Input The input contains at most 10 test cases. Each case contains two integers n and m (1 <= n <= 100,000, 1 <= m <= 109) on a single line. The input ends with n = m = 0. Output For each test case, print T(n) mod m. Sample Input 3 100 4 10 0 0 Sample Output 8 2
Binary Indexed Tree 是怎么做的
Problem Description Recently, Mr. Frog has learned binary indexed tree. Here is the code of adding t to the interval [1,x]: void add (int x, int t ){ for (int i = x; i != 0; i -= i & (-i)) a[i] += t; } If Mr. Frog is required to add t to the interval [l,r], he will add(r,t), and then add(l - 1,-t). The cost of an interval [l,r] is defined as the number of the “really changed point”. The “really changed point” is the point whose value is modified by the given code. For example, in order to add 1 to the interval [6,6], Mr. Frog will add 1 to the interval [1,6] (a[6] and a[4] will be added by 1), and add -1 to the interval [1,5] (a[5] and a[4] will be added by -1). As the result, a[6] will be added by 1, a[5] will be added by -1, and a[4] will be added by 0. a[6] and a[5] are “really changed point”, and the cost is 2. Mr. Frog wants to calculate the sum of the cost of the interval [l,r]⊆  [1,n] where l and r are two integers. Help Mr. Frog solve the problem. Input The first line contains only one integer T (T≤10000), which indicates the number of test cases. For each test case, it contains an integer n (1≤n≤1018). Output For each test case, output one line ”Case #x: y”, where x is the case number (starting from 1), y is the sum of the cost (modulo 1000000007). Sample Input 3 1 2 3 Sample Output Case #1: 1 Case #2: 4 Case #3: 10
如何将一个数组存入mysql
class a { byte[] c=new byte[5]; int[] d=new int[5]; } 怎么将该类寸入数组。 mysql中的表的对应的字段因该怎么设置。。。 谢谢各位好心人!!!!!!!!!!!1 [b]问题补充:[/b] binaray 对应java的string类型吗 [b]问题补充:[/b] 我的mysql数据库里 存了 表名为a的 一个表 表里面有两个binaray 字段 然后将用myeclipse 的反向工程,不过binaray对应的字段在java里面为string类型的 [b]问题补充:[/b] 非常感谢lovewhzlq , 我也有点疑惑的是,我以前用的一个插件在反向工程的时候 ,将binary的字段对应成了java的byte[]数组,可是在存的时候就是存不进去,但是又没有报异常
Counting Binary Trees
Problem Description There are 5 distinct binary trees of 3 nodes: Let T(n) be the number of distinct non-empty binary trees of no more than n nodes, your task is to calculate T(n) mod m. Input The input contains at most 10 test cases. Each case contains two integers n and m (1 <= n <= 100,000, 1 <= m <= 109) on a single line. The input ends with n = m = 0. Output For each test case, print T(n) mod m. Sample Input 3 100 4 10 0 0 Sample Output 8 2
The most binary substring 如何来编写呢
Problem Description Everybody knows the number is saved with the binary string in the computer. Now, their have N (1 <= N <= 1000) binary strings I tell you, your task is tell me what is the most binary substring with K (1 <= K <= L) characters in the strings I give you. The length of each string is L (1 <= L <= 60). Output times of the most appearing binary substring. Input Each line will contain three numbers N, L, K. Following N lines, represent N binary strings. Output One answer one line. Sample Input 2 6 3 101011 110101 2 6 4 101011 110101 Sample Output 4 2
c++用jsoncpp读取json文件并,有数千个数据,求把这些数据转成数组?
我有一个json文件,现在想把里面的数据读取出来,并转化为数组或是线性表。 文件是这样的,有数千个类似的这样的。我在网上找了很多的教程,但是里面的例子一般都是直接 int comment = root["comment"].asInt(); 但是我这里有数千个,而且后面要统计from相同的数量等等,所以想用数组或者是线性表来存储,方便后面。 因为是第一次接触json,完全菜鸟,所以很多概念都不了解,希望各位大神能够给一些简单一点的回答,非常感谢 [ { "comment": 0, "from": "北京-北京市-海淀区", "comments": [], "to": "河北省-承德市-隆化县", "score": "0", "corp": "兴铁物流", "time": "1398175318.988036" }, void readFileJson() { 根据网上的教程就写了一点点,我把添加到了源文件那里,可是好像打不开,运行的话就直接弹出Error poening file void readFileJson() { Json::Reader reader;//json解析 Json::Value root;//表示一个json格式的对象  ifstream in("rating(1).json", ios::binary); if (!in.is_open()) { cout << "Error opening file\n"; return; }
VB如何用二进制打开一个jpg 文件存到一个数组,每次存256循环存完。
硬盘已经有文件, Dim fileName As String ’定义了文件路径 Dim plainText() As Byte’用来存放照片二进制数据文件的数组 Open fileName For Binary As 1#’我已经写了用二进制打开文件 '现在循环就不知道怎么写了,想每次存256个字节到plainText()数组一直循环到整个照片全部存入数组。 For i=1 To FileLen(fileName)’但是这个是每次多一个,我要的是每次256的 求助啊 for i*256 =1 to ileLen(fileName)' 这样提示语句错误 拜托了
Byte Me! C语言的解法
Description You are a dealer at The One, the first all-binary casino in Las Vegas. What makes The One special is that its blackjack tables don't use cards; they use bytes (an 8-bit sequence representing a number from 0 to 255) and nibbles (a 4-bit sequence representing a number from 0 to 15). All day long, you play the house's hand against individual opponents. Of course, the casino owners know their statistics, and they have devised a strategy for you that gives gamblers just less than even odds. Here are the rules of binary blackjack: The goal of the game is to be the player closest to 510 points without going over. Each player is dealt two bytes, one face up and one face down. The players then have the opportunity to take more bytes (by saying, "Byte Me!") or more nibbles (by saying, "Nibble Me!") until he reaches his limit of 4 hits or has more than 510 points showing. All hits are played face up. If a player goes over 510, he immediately busts and loses the hand. The dealer hits last. The dealer wins any ties (this includes a tie where everyone busts). The rules for the dealer are (in order of precedence, where lower numbered rules override higher numbered ones): Never hit when it is certain that you've won by simply looking at your hand and what is showing of other people's hands. If your total is strictly less than 382 take a byte hit. If your total is less than or equal to 500 take a nibble hit. Take no hits Input Input to this problem will consist of a (non-empty) series of up to 100 data sets. Each data set will be formatted according to the following description, and there will be no blank lines separating data sets. A single data set has 6 components: Start line : A single line, "START N", where N is the number of players playing this hand (not including the dealer). There will never be more than 10 non-dealer players, and the dealer never plays by himself. Dealer Line : A single line containing 2 binary strings of exactly eight digits separated by exactly one space. These two strings represent the two cards in the dealer's hand. Player Line : A single line containing N 8-digit binary strings, each separated from the others by a single space. These represent the face-up cards of all of the non-dealer players. Byte Line : A single line containing 4 8-digit binary strings, each separated from the others by a single space. These represent the next 4 bytes in the byte deck, in the order they will be drawn. Nibble Line : A single line containing 4 4-digit binary strings, each separated from the others by a single space. These represent the next 4 nibbles in the nibble deck, in the order they will be drawn. End line : A single line, "END". Following the final data set will be a single line, "ENDOFINPUT". Here are some other useful facts: Oddly enough, each non-dealer player is always dealt a face-down card 11111111, value 255, but the dealer has no knowledge of this. Players other than the dealer never hit (they aren't too bright). Output Calculate the actions taken by the dealer and how the dealer fares with the resulting hand. For each data set, there will be exactly one output set, consisting of the following components: Hand Line : A single line, "HAND N", where N is the number of players playing this hand (not including the dealer). Dealer Hit List : A single line will be printed for each hit the dealer takes on his turn. For a byte hit, print a line "Byte me!", and for a nibble hit print, "Nibble me!" Outcome Line : A single line containing "Win!" if the dealer wins, "Bust!" if the dealer loses by busting, and "Lose!" if the dealer loses without busting. Sample Input START 1 11111111 11111111 00000001 10101010 01010101 11110000 00001111 1010 0101 1100 0011 END START 1 10111110 10111111 11111110 00010010 10101010 01010101 11110000 0001 1010 1100 0011 END START 8 11111111 00001000 00000000 00000001 00000010 00000011 00000100 00000101 00000110 00000111 00010010 10101010 01010101 11110000 0001 1010 1100 0011 END ENDOFINPUT Sample Output HAND 1 Win! HAND 1 Byte me! Nibble me! Nibble me! Nibble me! Lose! HAND 8 Win!
DenseFeature作为函数式API的第一层时报AttributeError: 'DenseFeatures' object has no attribute 'shape';该怎么解决啊。
我在用TensorFlow2.0搭建一个简单的全连接网络,第一层我设计的是一个DenseFeature,剩下的是三个Dense层,但我运行的时候却提示我AttributeError: 'DenseFeatures' object has no attribute 'shape'; 代码如下: ``` feature_layer = tf.keras.layers.DenseFeatures(one_order_feature_columns) dense1 = tf.keras.layers.Dense(128, activation='relu')(feature_layer) dense2 = tf.keras.layers.Dense(128, activation='relu')(dense1) dense3 = tf.keras.layers.Dense(1, activation='sigmoid')(dense2) model = tf.keras.Model(inputs=[feature_layer], outputs=dense3) # model = tf.keras.Sequential([ # tf.keras.layers.DenseFeatures(one_order_feature_columns), # tf.keras.layers.Dense(128, activation='relu'), # tf.keras.layers.Dense(128, activation='relu'), # tf.keras.layers.Dense(1, activation='sigmoid') # ]) model.compile(optimizer='adam', loss='binary_crossentropy', metrics=['accuracy']) model.fit(train_ds, epochs=5) ``` 我也尝试直接使用Sequential容器来搭建模型(代码中的注释部分),模型能够跑通。但使用函数式API时却不行。我是在哪出错了吗?
Skew Binary 有关二进制的计算
Description It had been a year since Swamp County Computing established a functional programming group. Your (non-functional programming) group is going to throw a surprise party for the anniversary. Now the functional folks really like skew binary numbers for some reason. "Easy to increment and decrement!" they say. Your task is to write a program to convert decimal integers to skew binary in the format they like. This will help in making banners and other party material. Number representations are made up of a list of digits. Call the lowest order digit the rank 0 digit, the next, rank 1, etc. For example, in decimal representation, digits are 0-9, the rank 0 digit has weight 1, the rank 1 digit has weight 10, and the rank i digit has weight 10i. In binary representation, the digits are 0 and 1, and the rank i digit has weight 2i. In skew binary representation, the digits are 0, 1, and 2, and the rank i digit has weight 2i+1 -1. Rank Weight 0 1 1 3 2 7 3 15 4 31 5 63 6 127 7 255 . . : : Allowing the digit 2 in the skew binary means there may be several ways to represent a given number. However the convention is that the digit 2 may only appear as the lowest rank non-zero digit. This makes the representation unique. In this problem, you should use a special way to write skew binary numbers as a list of ranks of non-zero digits in the number. The digit 2 is represented by the rank of the digit appearing twice in the list. Note that this means that only the first two ranks in the list may be equal. Each rank is a decimal integer, and is separated from the next rank by a comma (','). A list is started by a '[' and ended by a ']'. For example, the decimal number 5, which has the skew representation 12, should be written as [0,0,1]. Decimal 0 is an empty list: []. Input The first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by t lines, each containing a single decimal number with no leading or trailing white space. Each number will be in the range 0...100663270 (inclusive). Output There should be one line per test case containing the input decimal number, with no leading zeros or spaces, a single space, and the skew binary equivalent in list format with no leading or trailing spaces. Within the list each rank should have no extra leading zeros or leading or trailing spaces. Sample Input 5 0 1 2 3 4 Sample Output 0 [] 1 [0] 2 [0,0] 3 [1] 4 [0,1]
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