• Thinking_in_C++Annotated+Solution+Guide+for_Vol1

    Thinking_in_C++Annotated+Solution+Guide+for_Vol1 Thinking_in_C++课后题答案

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  • nim(简单)_组合数学

    Time limitation: 1 seconds Memory limitation: 64M The game of nim is played as follows. Some number of sticks are placed in a pile. Two players alternate in removing either one or two from the pile. The player who remove the last stick is the loser. The opponent remove sticks at the first, then it's your turn. Write a program to determine how many sticks should be removed when there's n sticks left in the pile. Input The first line is a integer n, which present how many data below. There's n integers m1, m2, ...mn on the following n lines, each line contains one of them, these number present how many sticks are there in the pile. Output For each mi, output how many sticks should be remove in this turn. If you will always lose no matter how many sticks are removed, output 0. Input Sample 3 4 5 6 Output Sample 0 1 2 Source BIT AsiaInfo CUP Programming Contest 2007

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  • nim_组合数学

    Description Nim is a 2-player game featuring several piles of stones. Players alternate turns, and on his/her turn, a player’s move consists of removing one or more stones from any single pile. Play ends when all the stones have been removed, at which point the last player to have moved is declared the winner. Given a position in Nim, your task is to determine how many winning moves there are in that position. A position in Nim is called “losing” if the first player to move from that position would lose if both sides played perfectly. A “winning move,” then, is a move that leaves the game in a losing position. There is a famous theorem that classifies all losing positions. Suppose a Nim position contains n piles having k1, k2, …, kn stones respectively; in such a position, there are k1 + k2 + … + kn possible moves. We write each ki in binary (base 2). Then, the Nim position is losing if and only if, among all the ki’s, there are an even number of 1’s in each digit position. In other words, the Nim position is losing if and only if the xor of the ki’s is 0. Consider the position with three piles given by k1 = 7, k2 = 11, and k3 = 13. In binary, these values are as follows: 111 1011 1101 There are an odd number of 1’s among the rightmost digits, so this position is not losing. However, suppose k3 were changed to be 12. Then, there would be exactly two 1’s in each digit position, and thus, the Nim position would become losing. Since a winning move is any move that leaves the game in a losing position, it follows that removing one stone from the third pile is a winning move when k1 = 7, k2 = 11, and k3 = 13. In fact, there are exactly three winning moves from this position: namely removing one stone from any of the three piles. Input The input test file will contain multiple test cases, each of which begins with a line indicating the number of piles, 1 ≤ n ≤ 1000. On the next line, there are n positive integers, 1 ≤ ki ≤ 1, 000, 000, 000, indicating the number of stones in each pile. The end-of-file is marked by a test case with n = 0 and should not be processed. Output For each test case, write a single line with an integer indicating the number of winning moves from the given Nim position. Sample Input 3 7 11 13 2 1000000000 1000000000 0 Sample Output 3 0 Source Stanford Local 2005

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  • 作业5 离散对数

    一、编程/分析作业 (1分) 使用C或C++实现EIGamal密码体制。 使用C或C++实现ECC EIGamal密码体制。 调用开源密码算法库,使用EIGamal加密一个字符串。 二、课后习题 (9分) 6.9(计算出3个密文对应的明文即可) 6.17 6.21 6.22 注: 编程题任选其一 课后习题任选其二。

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  • 作业4 RSA和大数因式分解

    一、编程/分析作业(1分) 使用C或C++实现RSA算法。提交源程序及测试结果。 调用开源密码算法库,使用RSA加密算法加密一个字符串。 调用开源密码算法库,使用Rabin加密算法加密一个字符串。 二、课后习题 5.5 5.7 5.9 5.14 5.16 5.17 5.23 注: 编程题任选其一 课后习题任选其二。

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  • 作业3 HASH & MAC

    一、课后习题 (9分) 4.6 4.9 4.10 4.11 4.12 注:(1,1)伪造者和(1,2)伪造者概念见书本109页 注:课后习题任选其二。 二、编程 (附加题 1分) 使用C或C++语言实现MD5算法。提交源程序。 使用C或C++语言实现SHA-1算法。提交源程序。 修改密码算法库HASH算法或者自己实现的HASH算法中HASH输出为原始输出的前8位,试分别求第一原像,第二原像以及碰撞。 注:编程作业任选其一。

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  • C++编程思想(第2版)_第2卷_实用编程技术

    C++编程思想(第2版)_第2卷_实用编程技术(中文版)

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