Mac下 charles4.0 抓不到电脑包 可以抓手机的

1个回答

Playfair Cipher 加密的问题
Problem Description The Playfair cipher is a manual symmetric encryption technique and was the first digraph substitution cipher. The scheme was invented in 1854 by Charles Wheatstone, but bears the name of Lord Playfair who promoted the use of the cipher. The Playfair cipher uses a 5 by 5 table containing each letter in the English alphabet exactly once (except 'Q' which is missing). The table constitutes the encryption key. To more easily remember the table, it is typically generated from a key phrase. First fill in the spaces in an empty table with the letters of the key phrase (dropping spaces and duplicate letters), then fill the remaining spaces with the rest of the letters of the alphabet in order. The key phrase is written in the top rows of the table, from left to right. For instance, if the key phrase is "playfair example", the encryption key becomes To encrypt a message, one would remove all spaces and then break the message into digraphs (groups of 2 letters) such that, for example, "Hello World" becomes "HE LL OW OR LD". Then map them out on the key table, and apply the rule below that matches the letter combination: If both letters are the same (or only one letter is left), add an 'X' after the first letter. Encrypt the new pair and continue (note that this changes all the remaining digraphs). If the letters appear on the same row of your table, replace them with the letters to their immediate right respectively (wrapping around to the left side of the row if a letter in the original pair was on the right side of the row). With the table above, the digraph 'CH' would be encrypted 'DB'. If the letters appear on the same column of your table, replace them with the letters immediately below respectively (wrapping around to the top side of the column if a letter in the original pair was on the bottom side of the column). With the table above, the digraph 'VA' would be encrypted 'AE'. If the letters are not on the same row or column, replace them with the letters on the same row respectively but at the other pair of corners of the rectangle defined by the original pair. The order is important { the first letter of the encrypted pair is the one that lies on the same row as the first letter of the plaintext pair. With the table above, the digraph 'KM' would be encrypted 'SR'. Write a program that reads a key phrase and a plaintext to encrypt, and outputs the encrypted text. The text to encrypt will not contain two 'x's following each other, or an 'x' as the last character, as this might cause the first rule above to repeat itself indefinitely. Input The input contains two lines. The first line contains the key phrase. The second line contains the text to encrypt. Each line will contain between 1 and 1000 characters, inclusive. Each character will be a lower case English letter, 'a' - 'z' (except 'q'), or a space character. Neither line will start or end with a space. Output The output should be a single line containing the encrypted text, in upper case. There should be no spaces in the output. Sample Input playfair example hide the gold in the tree stump the magic key i love programming competition Sample Output BMNDZBXDKYBEJVDMUIXMMNUVIF YDVHCWSPKNTAHKUBIPERMHGHDVRU
mac charles无法进行模拟弱网环境
Jenga 程序的编写
Problem Description In their spare time of training, Alice and Charles often play Jenga together. As they've played the game together so many times, they both know each others' performance as well as themselves'. Now with their success rate of each move provided, can you tell in what probability each of them will win? And of course, Alice and Charles, like other ACM-ICPC contestants such as you, are very clever. Jenga is a game of physical and mental skill. In Jenga, players take turns to remove a block from a tower and balance it on top, creating a taller and increasingly unstable structure as the game progresses. Jenga is played with 54 wooden blocks. Each block is three times as long as it is wide. To set up the game, the included loading tray is used to stack the initial tower which has 18 levels of three blocks placed adjacent to each other along their long side and perpendicular to the previous level (so, for example, if the blocks in the first level lie lengthwise north-south, the second level blocks will lie east-west). Once the tower is built, the players take turns to move. Moving in Jenga consists of taking one and only one block from any level (except those mentioned later) of the tower, and placing it on the topmost level in order to complete it. The blocks in the top level, and the level below it if the top level is not completed, cannot be removed. However, if the top level is completed, the blocks in the one below it can be removed. The removed block should be placed to make the top level as same as the other level (with no block removed). The move is successful if the tower does not fall. The game ends when the tower falls, or no block can be removed without making the tower fall (rarely happened). And the loser is the player who made the tower fall (i.e., whose turn it was when the tower fell), or who cannot make the move. Now let's consider each level of the tower, there're only four types of valid arrangement of wooden blocks, as illustrated above. At the beginning of the game, they're all of the type A (or rotated by 90 degrees). And by removing a block from type A, one will get either type B or type C (or the mirrored equivalent of type C). No block from type B can be removed without making the tower fall. From type C we can only remove a block and result in type D. Then no block can be removed further. So there are only three types of moves: (1) A -> B, (2) A -> C and (3) C -> D, in addition to adding the removed block to the top level. As Alice and Charles have played Jenga so many times, their success rate of each move is very stable and can be formulated as P = b - d*n, where b is the player's base success rate of this type of move, d is the decrease of success rate for each additional level, and n is the number of levels in the tower before this move. The incomplete top level also counts as one level. For example, if the game begins with 18 levels, and both players have the same performance with b = 2.8 and d = 0.1, then P will be 1.0 for the first turn, and become 0.9 between the 2nd and the 4th turns. If P does not lie in the range [0, 1], the nearest number in the range is indicated. (E.g. when a player cannot fail the first several moves, P will be more than 1 until n is a bit larger.) Input The input file contains multiple test cases. The first line of the input file is a single integer T (T ≤ 500), the number of test cases. Each test cases begins with a line of n0 (3 ≤ n0 ≤ 18), the number of levels in the tower when the game starts. (When n0 is not 18, the rule sare the same.) The second line contains 6 real numbers ba1, da1, ba2, da2, ba3, da3, indicating Alice's base success rates and the decreases of success rates for each of the three moves: (1) A -> B, (2) A -> C and (3) C -> D. The third line also contains 6 real numbers bc1, dc1, bc2, dc2, bc3, dc3, those of Charles. (0 ≤ b - d*n0 ≤ 2 and 0 < d ≤ 0.5 for all the 6 pairs of parameters. No real number will have more than 4 digits after the decimal point.) Output For each test case, print a line with Alice's winning probability, assume that she always moves first. Your answer should be rounded to the 4th digit after the decimal point. Sample Input 2 3 1.3 0.1 1.3 0.1 1.3 0.1 1.3 0.1 1.3 0.1 1.3 0.1 4 1.5 0.1 1.5 0.1 1.5 0.1 1.5 0.1 1.5 0.1 1.5 0.1 Sample Output 0.1810 0.8190
charles如何实现美团外卖抓包？

Jenga 是如何实现的呢
Problem Description In their spare time of training, Alice and Charles often play Jenga together. As they've played the game together so many times, they both know each others' performance as well as themselves'. Now with their success rate of each move provided, can you tell in what probability each of them will win? And of course, Alice and Charles, like other ACM-ICPC contestants such as you, are very clever. Jenga is a game of physical and mental skill. In Jenga, players take turns to remove a block from a tower and balance it on top, creating a taller and increasingly unstable structure as the game progresses. Jenga is played with 54 wooden blocks. Each block is three times as long as it is wide. To set up the game, the included loading tray is used to stack the initial tower which has 18 levels of three blocks placed adjacent to each other along their long side and perpendicular to the previous level (so, for example, if the blocks in the first level lie lengthwise north-south, the second level blocks will lie east-west). Once the tower is built, the players take turns to move. Moving in Jenga consists of taking one and only one block from any level (except those mentioned later) of the tower, and placing it on the topmost level in order to complete it. The blocks in the top level, and the level below it if the top level is not completed, cannot be removed. However, if the top level is completed, the blocks in the one below it can be removed. The removed block should be placed to make the top level as same as the other level (with no block removed). The move is successful if the tower does not fall. The game ends when the tower falls, or no block can be removed without making the tower fall (rarely happened). And the loser is the player who made the tower fall (i.e., whose turn it was when the tower fell), or who cannot make the move. Now let's consider each level of the tower, there're only four types of valid arrangement of wooden blocks, as illustrated above. At the beginning of the game, they're all of the type A (or rotated by 90 degrees). And by removing a block from type A, one will get either type B or type C (or the mirrored equivalent of type C). No block from type B can be removed without making the tower fall. From type C we can only remove a block and result in type D. Then no block can be removed further. So there are only three types of moves: (1) A -> B, (2) A -> C and (3) C -> D, in addition to adding the removed block to the top level. As Alice and Charles have played Jenga so many times, their success rate of each move is very stable and can be formulated as P = b - d*n, where b is the player's base success rate of this type of move, d is the decrease of success rate for each additional level, and n is the number of levels in the tower before this move. The incomplete top level also counts as one level. For example, if the game begins with 18 levels, and both players have the same performance with b = 2.8 and d = 0.1, then P will be 1.0 for the first turn, and become 0.9 between the 2nd and the 4th turns. If P does not lie in the range [0, 1], the nearest number in the range is indicated. (E.g. when a player cannot fail the first several moves, P will be more than 1 until n is a bit larger.) Input The input file contains multiple test cases. The first line of the input file is a single integer T (T ≤ 500), the number of test cases. Each test cases begins with a line of n0 (3 ≤ n0 ≤ 18), the number of levels in the tower when the game starts. (When n0 is not 18, the rule sare the same.) The second line contains 6 real numbers ba1, da1, ba2, da2, ba3, da3, indicating Alice's base success rates and the decreases of success rates for each of the three moves: (1) A -> B, (2) A -> C and (3) C -> D. The third line also contains 6 real numbers bc1, dc1, bc2, dc2, bc3, dc3, those of Charles. (0 ≤ b - d*n0 ≤ 2 and 0 < d ≤ 0.5 for all the 6 pairs of parameters. No real number will have more than 4 digits after the decimal point.) Output For each test case, print a line with Alice's winning probability, assume that she always moves first. Your answer should be rounded to the 4th digit after the decimal point. Sample Input 2 3 1.3 0.1 1.3 0.1 1.3 0.1 1.3 0.1 1.3 0.1 1.3 0.1 4 1.5 0.1 1.5 0.1 1.5 0.1 1.5 0.1 1.5 0.1 1.5 0.1 Sample Output 0.1810 0.8190

Charles这个软件，突然就不能抓取localhost和127.0.0.1的接口了

Game Simulator 模拟器的问题
Polish notation 程序的编写
Problem Description Reverse Polish notation (RPN) is a method for representing expressions in which the operator symbol is placed after the arguments being operated on. Polish notation, in which the operator comes before the operands, was invented in the 1920s by the Polish mathematician Jan Lucasiewicz. In the late 1950s, Australian philosopher and computer scientist Charles L. Hamblin suggested placing the operator after the operands and hence created reverse polish notation. RPN has the property that brackets are not required to represent the order of evaluation or grouping of the terms. RPN expressions are simply evaluated from left to right and this greatly simplifies the computation of the expression within computer programs. As an example, the arithmetic expression (3+4)*5 can be expressed in RPN as 3 4 + 5 *. Reverse Polish notation, also known as postfix notation, contrasts with the infix notation of standard arithmetic expressions in which the operator symbol appears between the operands. So Polish notation just as prefix notation. Now, give you a string of standard arithmetic expressions, please tell me the Polish notation and the value of expressions. Input There're have multi-case. Every case put in one line, the expressions just contain some positive integers(all less than 100, the number of integers less than 20), bi-operand operators(only have 3 kinds : +,-,*) and some brackets'(',')'. you can assume the expressions was valid. Output Each case output the Polish notation in first line, and the result of expressions was output in second line. all of the answers are no any spaces and blank line.the answer will be not exceed the 64-signed integer. Sample Input 1+2-3*(4-5) 1+2*(3-4)-5*6 Sample Output Case 1: - + 1 2 * 3 - 4 5 6 Case 2: - + 1 * 2 - 3 4 * 5 6 -31
Charles配置完毕后，抓取手机APP数据还是unknown
Jenga 求问如何解决
Problem Description In their spare time of training, Alice and Charles often play Jenga together. As they've played the game together so many times, they both know each others' performance as well as themselves'. Now with their success rate of each move provided, can you tell in what probability each of them will win? And of course, Alice and Charles, like other ACM-ICPC contestants such as you, are very clever. Jenga is a game of physical and mental skill. In Jenga, players take turns to remove a block from a tower and balance it on top, creating a taller and increasingly unstable structure as the game progresses. Jenga is played with 54 wooden blocks. Each block is three times as long as it is wide. To set up the game, the included loading tray is used to stack the initial tower which has 18 levels of three blocks placed adjacent to each other along their long side and perpendicular to the previous level (so, for example, if the blocks in the first level lie lengthwise north-south, the second level blocks will lie east-west). Once the tower is built, the players take turns to move. Moving in Jenga consists of taking one and only one block from any level (except those mentioned later) of the tower, and placing it on the topmost level in order to complete it. The blocks in the top level, and the level below it if the top level is not completed, cannot be removed. However, if the top level is completed, the blocks in the one below it can be removed. The removed block should be placed to make the top level as same as the other level (with no block removed). The move is successful if the tower does not fall. The game ends when the tower falls, or no block can be removed without making the tower fall (rarely happened). And the loser is the player who made the tower fall (i.e., whose turn it was when the tower fell), or who cannot make the move. Now let's consider each level of the tower, there're only four types of valid arrangement of wooden blocks, as illustrated above. At the beginning of the game, they're all of the type A (or rotated by 90 degrees). And by removing a block from type A, one will get either type B or type C (or the mirrored equivalent of type C). No block from type B can be removed without making the tower fall. From type C we can only remove a block and result in type D. Then no block can be removed further. So there are only three types of moves: (1) A -> B, (2) A -> C and (3) C -> D, in addition to adding the removed block to the top level. As Alice and Charles have played Jenga so many times, their success rate of each move is very stable and can be formulated as P = b - d*n, where b is the player's base success rate of this type of move, d is the decrease of success rate for each additional level, and n is the number of levels in the tower before this move. The incomplete top level also counts as one level. For example, if the game begins with 18 levels, and both players have the same performance with b = 2.8 and d = 0.1, then P will be 1.0 for the first turn, and become 0.9 between the 2nd and the 4th turns. If P does not lie in the range [0, 1], the nearest number in the range is indicated. (E.g. when a player cannot fail the first several moves, P will be more than 1 until n is a bit larger.) Input The input file contains multiple test cases. The first line of the input file is a single integer T (T ≤ 500), the number of test cases. Each test cases begins with a line of n0 (3 ≤ n0 ≤ 18), the number of levels in the tower when the game starts. (When n0 is not 18, the rule sare the same.) The second line contains 6 real numbers ba1, da1, ba2, da2, ba3, da3, indicating Alice's base success rates and the decreases of success rates for each of the three moves: (1) A -> B, (2) A -> C and (3) C -> D. The third line also contains 6 real numbers bc1, dc1, bc2, dc2, bc3, dc3, those of Charles. (0 ≤ b - d*n0 ≤ 2 and 0 < d ≤ 0.5 for all the 6 pairs of parameters. No real number will have more than 4 digits after the decimal point.) Output For each test case, print a line with Alice's winning probability, assume that she always moves first. Your answer should be rounded to the 4th digit after the decimal point. Sample Input 2 3 1.3 0.1 1.3 0.1 1.3 0.1 1.3 0.1 1.3 0.1 1.3 0.1 4 1.5 0.1 1.5 0.1 1.5 0.1 1.5 0.1 1.5 0.1 1.5 0.1 Sample Output 0.1810 0.8190
knockout.js 关于用于嵌套的数组无关系的foreach多层嵌套报undefined的错误
# 问题描述： 我要实现ko的foreach多层嵌套，用于foreach的两个数组没有关系，第二层foreach绑定的数组报错： Message: Unable to process binding "foreach: function (){return p2 }" Message: p2 is not defined # 代码： js: viewModel = { p1 : ko.observableArray( [ { firstName: 'Bert', lastName: 'Bertington' }, { firstName: 'Charles', lastName: 'Charlesforth' }, { firstName: 'Denise', lastName: 'Dentiste' } ] ), p2 : ko.observableArray( [ { firstName: 'Bert', lastName: 'Bertington' }, { firstName: 'Charles', lastName: 'Charlesforth' }, { firstName: 'Denise', lastName: 'Dentiste' } ] ), } ``` html: <div data-bind="foreach:p1"> <div>我是p1</div> <div data-bind="foreach:p2"> <li>我是p2</li> </div> </div> 报错： knockout-3.2.0.debug.js:2890 Uncaught ReferenceError: Unable to process binding "foreach: function (){return p1 }" Message: Unable to process binding "foreach: function (){return p2 }" Message: p2 is not defined at foreach (eval at createBindingsStringEvaluator (knockout-3.2.0.debug.js:2503), <anonymous>:3:60) at knockout-3.2.0.debug.js:3739 at Object.init (knockout-3.2.0.debug.js:4859) at init (knockout-3.2.0.debug.js:3764) at knockout-3.2.0.debug.js:2867 at Object.ignore (knockout-3.2.0.debug.js:1194) at knockout-3.2.0.debug.js:2866 at Object.arrayForEach (knockout-3.2.0.debug.js:121) at applyBindingsToNodeInternal (knockout-3.2.0.debug.js:2852) at applyBindingsToNodeAndDescendantsInternal (knockout-3.2.0.debug.js:2732) 请各位大牛帮忙看看。 谢谢。
iOS请求https接口，为什么用charles或者fiddler等抓包工具截取的数据是明文的？

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