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 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…

// 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…

题目链接:hdu_5831_Rikka with Parenthesis II 题意: 给你一些括号的排列,必须交换一次,能不能将全部的括号匹配 题解: 模拟一下括号的匹配就行了,注意要特判只有一对括号是NO,因为必须要交换一次 #include <cstdio> #include <cstring> ; char s[N]; int st[N],top; int main() { int t; scanf("%d", &t); while (t--)…

数字的反转: 就是将数字倒着存下来而已.(*^__^*) 嘻嘻…… 大致思路:将数字一位一位取出来,存在一个数组里面,然后再将其变成数字,输出. 详见代码. while (a) //将每位数字取出来,取完为止 { num1[i]=a%; //将每一个各位取出存在数组里面,实现了将数字反转 i++; //数组的变化 a/=; } 趁热打铁 例题:hdu 4554 叛逆的小明 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4554 叛逆的小明 Time…

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…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5265 pog loves szh II Description Pog and Szh are playing games.There is a sequence with n numbers, Pog will choose a number A from the sequence. Szh will choose an another number named B from the rest i…

pog loves szh II Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5265 Description pog在与szh玩游戏,首先pog找到了一个包含n个数的序列,然后他在这n个数中挑出了一个数A,szh出于对pog的爱,在余下的n−1个数中也挑了一个数B,那么szh与pog的恩爱值为(A+B)对p取模后的余数,pog与szh当然想让恩爱值越高越好,并且他们…

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…

开始学习最小树形图,模板题. Ice_cream’s world II Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) [Problem Description] After awarded lands to ACMers, the queen want to choose a city be her capital. This is an important event in…

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…

A + B Problem II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 347161    Accepted Submission(s): 67385 Problem Description I have a very simple problem for you. Given two integers A and B, you…

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>…

- Eddy's research II Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description As is known, Ackermann function plays an important role in the sphere of theoretical computer science. However, in the other h…