31 Rikka with Parenthesis II (六花与括号II) Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Description 题目描述 As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to…

Rikka with Parenthesis II 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5831 Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Correct parenthe…

Rikka with Parenthesis II 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5831 Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Correct parenthe…

Rikka with Parenthesis II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situat…

Rikka with Parenthesis II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 136    Accepted Submission(s): 97 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this sit…

题目:传送门. 题意:T组数据,每组给定一个长度n,随后给定一个长度为n的字符串,字符串只包含'('或')',随后交换其中两个位置,必须交换一次也只能交换一次,问能否构成一个合法的括号匹配,就是()()或者((()))这种的. 题解:首先n为奇数肯定是No,左括号和右括号个数不相等是No,n=2的时候如果是()也是no,因为必须交换一次,就会变成)(,所以是No.否则如果出现一个没有与其相匹配的右括号,就是右括号出现在与他匹配的左括号之前,如果这种情况出现了三次或三次以上就是No,其余是Yes.…

如果左括号数量和右括号数量不等,输出No 进行一次匹配,看匹配完之后栈中还有多少元素: 如果n=2,并且栈中无元素,说明是()的情况,输出No 如果n=2,并且栈中有元素,说明是)(的情况,输出Yes 如果n>2,并且栈中没有元素,或者有2个,或者有4个,输出Yes 如果n>2,并且栈中元素个数大于4个,输出No #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #i…

用一个temp变量,每次出现左括号,+1,右括号,-1:用ans来记录出现的最小的值,很显然最终temp不等于0或者ans比-2小都是不可以的.-2是可以的,因为:“))((”可以把最左边的和最右边的交换即可,其他-2的情形同理.另外要注意的坑点是Hint里面所说的:“But do nothing is not allowed.”.因此,“()”是不可以的,这个要特判. 代码如下: #include <stdio.h> #include <algorithm> #include &…

BUPT2017 wintertraining(16) #4 G HDU - 5831 题意 给定括号序列,问能否交换一对括号使得括号合法. 题解 注意()是No的情况. 任意时刻)不能比(超过2个以上. 最后)和(的差距要在两个以内,且n必须是偶数. 代码 #include <cstdio> #include <cstring> #include <algorithm> #include <iostream> using namespace std; ch…

Rikka with sequence Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5204 Description 众所周知,萌萌哒六花不擅长数学,所以勇太给了她一些数学问题做练习,其中有一道是这样的:现在有一个序列,因为这个序列很任性,开始时空的.接下来发生了n个事件,每一个事件是以下两种之一:1.勇太利用黑炎龙的力量在序列的开头.结尾以及每相邻两个元素之间都插入…

Rikka with Graph 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5631 Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has a non-direct gra…

Rikka with Tree  Accepts: 207  Submissions: 815  Time Limit: 2000/1000 MS (Java/Others)  Memory Limit: 65536/65536 K (Java/Others) 问题描述 众所周知,萌萌哒六花不擅长数学,所以勇太给了她一些数学问题做练习,其中有一道是这样的: 对于一棵树TT,令F(T,i)F(T,i)为点1到点ii的最短距离(边长是1). 两棵树AA和BB是相似的当且仅当他们顶点数相同且对于任意的…

// 判断相同区间(lazy) 多校8 HDU 5828 Rikka with Sequence // 题意:三种操作,1增加值,2开根,3求和 // 思路:这题与HDU 4027 和HDU 5634 差不多 // 注意开根号的话,遇到极差等于1的,开根号以后有可能还是差1.如 // 2 3 2 3... // 8 9 8 9... // 2 3 2 3... // 8 9 8 9... // 剩下就是遇到区间相等的话,就直接开根号不往下传 #include <bits/stdc++.h> u…

思路来自 某FXXL 不过复杂度咋算的.. /* HDU 6091 - Rikka with Match [ 树形DP ] | 2017 Multi-University Training Contest 5 题意: 给出N个点的树,求去边的方案数使得 去边后最大匹配数是M的倍数 限制: N<=5e4, M<=200 分析: 设 DP[u][i][0] 表示 以点 u 为根的子树 最大匹配数模 m 为 i 时,且 u 点没有匹配的方案数 DP[u][i][1] 表示 以点 u 为根的子树 最大…

思路和任意模数FFT模板都来自 这里 看了一晚上那篇<再探快速傅里叶变换>还是懵得不行,可能水平还没到- - 只能先存个模板了,这题单模数NTT跑了5.9s,没敢写三模数NTT,可能姿势太差了... 具体推导大概这样就可以了: /* HDU 6088 - Rikka with Rock-paper-scissors [ 任意模数FFT,数论 ] | 2017 Multi-University Training Contest 5 题意: 计算 3^n * ∑ [0<=i+j<=n]…

JAVA+大数搞了一遍- - 不是很麻烦- - /* HDU 6093 - Rikka with Number [ 进制转换,康托展开,大数 ] | 2017 Multi-University Training Contest 5 题意: 求L,R之间的好数的个数,好数要求在某个d(>=2)进制下数位是0到d-1的 分析: d 进制下好数的个数为 d!-(d-1)! ,且满足 d^(d-1) <= K <= d^d 可知 若 N > d^d 则 1-N 包含前 d-1 个进制的所有…

看了标程的压位,才知道压位也能很容易写- - /* HDU 6085 - Rikka with Candies [ 压位 ] | 2017 Multi-University Training Contest 5 题意: 给定 A[N], B[N], Q 个 k 对于每个k, 求 A[i] % B[j] == k 的 (i,j)对数 限制 N, Q <=50000 分析: 对于每个 B[i] 按其倍数分块,则对于 A[j] ∈ [x*B[i],(x+1)*B[i]) , A[j]%B[i] = A…

SOPC开发流程之NIOS II 处理器运行 UC/OS II 这里以在芯航线FPGA学习套件的核心板上搭建 NIOS II 软核并运行 UCOS II操作系统为例介绍SOPC的开发流程. 第一步:建立 Quartus II 工程 建立 Quartus II 工程时需要注意以下几点 1. 器件选择为 EP4CE10F17C8N: 2. 工程路径中不得出现非法字符(空格和中文字符): 3. 开发工具选择 Quartus II 11.0及以上,这里我选择的版本为 Quartus II 15.1. 4…

[063-Unique Paths II(唯一路径问题II)] [LeetCode-面试算法经典-Java实现][全部题目文件夹索引] 原题 Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How many unique paths would there be? An obstacle and empty space is marked as 1 and 0 re…

[107-Binary Tree Level Order Traversal II(二叉树层序遍历II)] [LeetCode-面试算法经典-Java实现][全部题目文件夹索引] 原题 Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example…

开发Altera Nios II软件可使用Nios II IDE或 Nios II Software BuildTools for Eclipse(即:Nios II SBT for Eclipse),使用“Build All”或“Build Project”编译工程后有时会出现如下错误:c:/altera/11.0/nios2eds/bin/gnu/h-i686-mingw32/bin/../lib/gcc/nios2-elf/4.1.2/../../../../nios2-elf/bin/l…

Rikka with Graph II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1051    Accepted Submission(s): 266 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situati…

http://acm.hdu.edu.cn/showproblem.php?pid=5424 哈密顿通路:联通的图,访问每个顶点的路径且只访问一次 n个点n条边 n个顶点有n - 1条边,最后一条边的连接情况: (1)自环(这里不需要考虑): (2)最后一条边将首和尾连接,这样每个点的度都为2: (3)最后一条边将首和除尾之外的点连接或将尾和出尾之外的点连接,这样相应的首或尾的度最小,度为1: (4)最后一条边将首和尾除外的两个点连接,这样就有两个点的度最小,度都为1 如果所给的图是联通的话,那…

Problem Description   As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has a non-direct graph with n vertices and n edges. Now he wants you to tell him…

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Correct parentheses sequences can be defined recursively as follows: 1.The empty string "" is a correc…

题目大意: 在 N 个点 N 条边组成的图中判断是否存在汉密尔顿路径. 思路:忽略重边与自回路,先判断是否连通,否则输出"NO",DFS搜索是否存在汉密尔顿路径. #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<queue> #include<algorithm> #include<cmath&g…

Rikka with Sequence 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5828 Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has an array A wi…

Rikka with Subset Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1846    Accepted Submission(s): 896 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation…

Rikka with Sequence 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5828 Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has an array A wi…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5831 给你一个只包含‘(’‘)’的字符串,然后让我们交换两个字符一次,问是否能得到一个合法的匹配:必须要交换一次,而且只能交换一次: () No ))(( Yes这两个是比较特殊的注意一下: 可以把串中的第一个')'和最后一个'('交换一下,然后判断是否合法即可: #include<iostream> #include<stdio.h> #include<string.h>…