题目大意:关于大数的mod和进制转换,直接使用Java的BigInteger类,正在copy式学习中... import java.io.*; import java.util.*; import java.math.*; class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); while (true) { int base = sc.nextInt(); if (bas…

Basic remains Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5221   Accepted: 2203 Description Given a base b and two non-negative base b integers p and m, compute p mod m and print the result as a base b integer. p mod m is defined as…

Basic remains input:   b p m    读入p进制的p,m,   求p%m   ,以b进制输出 1 import java.util.*; 2 import java.math.*; 3 import java.io.*; 4 import java.text.*; 5 6 public class Main 7 { 8 public static void main(String[] args) 9 { 10 // 对于大量输入,下面方式可能会快一些. 11 Scann…

Problem F: Polynomial Remains Given the polynomial a(x) = an xn + ... + a1 x + a0, compute the remainder r(x) when a(x) is divided by xk+1. The input consists of a number of cases. The first line of each case specifies the two integers n and k (0 ≤ n…

题意:给定n个只有大写字母组成的字符串,选取尽可能多的字符串,使得这些字符串中每个字母的个数都是偶数.n<=24 思路:直接枚举每个字符串的选或不选,复杂度是O(2^n).其实还有更简便的方法. 对于每个字母,其实具体出现了多少次并不重要,重要的是奇数次还是偶数次,我们用0对应奇数次,1对应偶数次.对于每个字符串,我们就可以计算出对应的二进制数,方法如下.如果A出现奇数次,那么二进制数第一个位置为1,偶数次为0:如果B出现奇数次,那么二进制数第二个位置为1,偶数次为0……以此类推,每个位置都有一…

题目链接:http://poj.org/problem?id=2305 ime Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5326   Accepted: 2267 Description Given a base b and two non-negative base b integers p and m, compute p mod m and print the result as a base b integer. p…

题解 在一个字符串中,每个字符出现的次数本身是无关紧要的,重要的只是这些次数的奇偶性,因此想到用一个二进制的位表示一个字母($1$表示出现奇数次,$0$表示出现偶数次).比如样例的$6$个数,写成二进制后如图所示. 此时,问题转化为求尽量多的数,使得它们的$xor$值为$0$. 最容易想到的方法是直接穷举,时间复杂度为$O(2^n)$,有些偏大.注意到$xor$值为$0$的两个整数必须完全相等,我们可以把字符串分成两个部分:首先计算前$n \over 2$个字符串所能得到的所有$xor$值,并将…

题意 给出B(10以内大于0)进制下 p (1000位以内)和m(9位以内) 求 p%m 在b进制下等于什么 思路: 可以计算   1e9不会溢出Int所以m在int值以内  先求m  要处理p  每次取一位模刚刚的m即可(b^n 也要不停模) 坑点 : p可能等于0  这时候要输出0 #include<cstdio> #include<cstring> #include<vector> #include<cmath> #include<iostrea…

训练指南p.59 #include <cstdio> #include <cstring> #include <cstdlib> #include <map> using namespace std; ; int N; int A[MAXN]; ]; map<int , int> table; int bitcount( int x ) { ; while ( x ) { ) ++res; x >>= ; } return res;…

1.Basic remains 题目链接 涉及内容: |大数读入|大数模|大数进制读入时转化为十进制|大数输出时转化为其他进制输出| import java.io.*; import java.math.*; import java.util.*; import java.text.*;//各种头文件 public class Main { public static void main(String[] args) { Scanner cin = new Scanner(System.in);…

题目大意:给一个正整数n,求出在[1, n]区间内和n互质的正整数的个数.Euler's Totient(欧拉函数)的直接应用. #include <cstdio> #include <vector> #include <algorithm> #include <bitset> using namespace std; typedef vector<int> vi; typedef long long ll; #define MAXN 10000…

题目 https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=966 题意 n个大写字母串(n <= 24),问最多取多少个,使得所有字符出现的次数都是偶数. 思路 如刘书 1. 由于最多出现26个字母,而且字母在字符串中出现的次数本身不重要,只要记录奇偶性,所以可以将这些字符串转化为01串便于存储. 2. 问题转化为最多取多少…

UVa 10387- Billiard Table of Contents 1 题目 2 思路 3 代码 4 参考 1 题目 ============= Problem A: Billiard In a billiard table with horizontal side  a  inches and vertical side  b  inches, a ball is launched from the middle of the table. After  s  > 0 seconds…

UVa 1262  Password 题目: Password   Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description  Shoulder-surfing is the behavior of intentionally and stealthily watching the screen of another person's electronic…

UVa 10618 Tango Tango Insurrection 题目: Problem A: Tango Tango Insurrection You are attempting to learn to play a simple arcade dancing game. The game has 4 arrows set into a pad: Up, Left, Down, Right. While a song plays, you watch arrows rise on a s…

UVA 11774 - Doom's Day 题目链接 题意:给定一个3^n*3^m的矩阵,要求每次按行优先取出,按列优先放回,问几次能回复原状 思路:没想到怎么推理,找规律答案是(n + m) / gcd(n, m),在topcoder上看到一个证明,例如以下: We can associate at each cell a base 3-number, the log3(R) most significant digits is the index of the row of the cel…

题目大意:Euler's Totient的应用. 几乎和UVa 10179 - Irreducable Basic Fractions一样,于是偷了个懒,直接用10179题的代码,结果WA了,感觉一样啊...然后就搜,看到n=1的特殊情况,好吧,读题的时候也小小注意了一下"less than" 呢,写代码时就忘的一干二净了...读题时应该把注意事项记下来的... #include <cstdio> #include <vector> #include <a…

UVA 1394 And Then There Was One / Gym 101415A And Then There Was One / UVAlive 3882 And Then There Was One / POJ 3517 And Then There Was One / Aizu 1275 And Then There Was One (动态规划,思维题) Description Let's play a stone removing game. Initially, n ston…

排版有点乱.... ========================================================================= Operational Amplifiers (OAs) are highly stable, high gain dc difference amplifiers. Sincethere is no capacitive coupling between their various amplifying stages, they…

Calendar 常数 可在代码中的任何地方用下列常数代替实际值: 常数 值 描述 vbCalGreg 0 指出使用的是阳历. vbCalHijri 1 指出使用的是伊斯兰历法. Color 常数 可在代码中的任何地方用下列常数代替实际值: 常数 值 描述 vbBlack 0x0 黑色 vbRed 0xFF 红色 vbGreen 0xFF00 绿色 vbYellow 0xFFFF 黄色 vbBlue 0xFF0000 蓝色 vbMagenta 0xFF00FF 紫红色 vbCyan 0xFFFF…

uva 10494 - If We Were a Child Again If We Were a Child Again Input: standard inputOutput: standard output Time Limit: 7 seconds   “Oooooooooooooooh! If I could do the easy mathematics like my school days!! I can guarantee, that I’d not make any mist…

Password Time Limit: 3000ms Memory Limit: 131072KB This problem will be judged on UVA. Original ID: 1262 64-bit integer IO format: %lld      Java class name: Main pid=18400" class="btn" style="">Prev Submit showpid=18401"…

The basic principle of test case evaluation is that output resulting from running a test case is compared to the expected result. https://dev.mysql.com/doc/dev/mysql-server/latest/PAGE_QUICK_START.html https://github.com/mysql/mysql-server/blob/5.7/m…

先上题目: F. Dancing the Cheeky-Cheeky  Context The Cheeky-Cheeky is a new song. They dance it in Mula, and also in Hong Kong. All the freekies dance it, and the geek all love it. And the Cheeky-Cheeky is danced like this: The breikin-brokin. The meneito…

Atitit HTTP认证机制基本验证 (Basic Authentication) 和摘要验证 (Digest Authentication)attilax总结 1.1. 最广泛使用的是基本验证 (Basic Authentication) 和摘要验证 (Digest Authentication).1 1.2. 关于HTTP AUTH的文档不多.1 1.3. 什么是HTTP基本认证1 1.4. 适用场合 路由器 摄像头2 1.5. 其他认证  除了基本认证(Basic Authenticat…

After learning the basic opreation of Redis,we should take some time to summarize the usage. And I wrote my first edition RedisHelper.Here is the code: The Interface IRedis: public interface IRedis { ITransaction GetTransaction(, bool isRead = false)…

Data play an important part in our project,how can we ensure correctness of the data and prevent the data from error.In relational database, we are famillar with the usage of transaction. begin opreations commit/rollback But there are some difference…

This post is mainly about the publishment and subscription in Redis.I think you may subscribe some offiial accounts on wechat,and when the authors published something new to their accounts,you will get them in your wechat.The instance can make you un…

Redis's List is different from C#'s List,but similar with C#'s LinkedList.Sometimes I confuse with them.I expect that you won't mix them and have a clear mind of them. There are 17 commands we can use in List. Push and pop are the base opreation of t…

The last post is mainly about the unsorted set,in this post I will show you the sorted set playing an important role in Redis.There are many command added after the version 2.8.9.OK,let's see the below picture firstly.There are 24 commands to handle…