Poj OpenJudge 1068 Parencodings】的更多相关文章

1.Link: http://poj.org/problem?id=1068 http://bailian.openjudge.cn/practice/1068 2.Content: Parencodings Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 20077   Accepted: 12122 Description Let S = s1 s2...s2n be a well-formed string of p…
题目地址:http://poj.org/problem?id=1068 /* 题意:给出每个右括号前的左括号总数(P序列),输出每对括号里的(包括自身)右括号总数(W序列) 模拟题:无算法,s数组把左括号记为-1,右括号记为1,然后对于每个右括号,numl记录之前的左括号 numr记录之前的右括号,sum累加s[i]的值,当sum == 0,并且s[i]是左括号(一对括号)结束,记录b[]记录numl的值即为答案 我当时题目没读懂,浪费很长时间.另外,可以用vector存储括号,还原字符串本来的…
链接: http://poj.org/problem?id=1068 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=27454#problem/B Parencodings Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 17044   Accepted: 10199 Description Let S = s1 s2...s2n be a well-forme…
http://poj.org/problem?id=1068 #include<cstdio> #include <cstring> using namespace std; int ind[45]; bool used[45]; int r[21]; int l[21]; int len,n,llen; int w[21]; int main(){ int t; scanf("%d",&t); while(t--){ memset(used,0,siz…
题目链接:http://poj.org/problem?id=1068 思路分析:对栈的模拟,将栈中元素视为广义表,如 (((()()()))),可以看做 LS =< a1, a2..., a12 >,对于可以配对的序列,如 <a4, a5>看做一个元素,其 W 值为1: 同理,<a6, a7>为一个元素,其W值为1,< a3, a4, a5, a6, a7, a8, a9, a10 >看做一个元素, 其W值为 <a4, a5> 与 <a6…
                                                                                                Parencodings Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 19550   Accepted: 11804 Description Let S = s1 s2...s2n be a well-formed string…
转载请注明出处:viewmode=contents">http://blog.csdn.net/u012860063?viewmode=contents 题目链接:http://poj.org/problem? id=1068 Description Let S = s1 s2...s2n be a well-formed string of parentheses. S can be encoded in two different ways:  q By an integer sequ…
Parencodings Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 24932   Accepted: 14695 Description Let S = s1 s2...s2n be a well-formed string of parentheses. S can be encoded in two different ways: q By an integer sequence P = p1 p2...pn…
Parencodings Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 19352   Accepted: 11675 Description Let S = s1 s2...s2n be a well-formed string of parentheses. S can be encoded in two different ways:  q By an integer sequence P = p1 p2...pn…
进入每个' )  '多少前' (  ', 我们力求在每' ) '多少前' )  ', 我的方法是最原始的图还原出来,去寻找')'. 用. . #include<stdio.h> #include<string.h> int y[505],t[505]; char s[505]; int main() { int a,b,i,j,u; scanf("%d",&a); while(a--) { memset(y,0,sizeof(y)); memset(t,…
Description Let S = s1 s2...s2n be a well-formed string of parentheses. S can be encoded in two different ways: q By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence). q B…
1.Link: http://poj.org/problem?id=1062 http://bailian.openjudge.cn/practice/1062/ 2.Content: 昂贵的聘礼 Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 37631   Accepted: 10872 Description 年轻的探险家来到了一个印第安部落里.在那里他和酋长的女儿相爱了,于是便向酋长去求亲.酋长要他用10000个金…
1.Link: http://poj.org/problem?id=1860 http://bailian.openjudge.cn/practice/1860 2.Content: Currency Exchange Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 20706   Accepted: 7428 Description Several currency exchange points are working…
1.Link: http://poj.org/problem?id=2602 http://bailian.openjudge.cn/practice/2602/ 2.Content: Superlong sums Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 22337   Accepted: 6577 Description The creators of a new programming language D++…
1.Link: http://poj.org/problem?id=2389 http://bailian.openjudge.cn/practice/2389/ 2.Content: Bull Math Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13067   Accepted: 6736 Description Bulls are so much better at math than the cows. The…
1.Link: http://poj.org/problem?id=1573 http://bailian.openjudge.cn/practice/1573/ 2.Content: Robot Motion Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 10856   Accepted: 5260 Description A robot has been programmed to follow the instru…
1.Link: http://poj.org/problem?id=2632 http://bailian.openjudge.cn/practice/2632/ 2.Content: Crashing Robots Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7912   Accepted: 3441 Description In a modernized warehouse, robots are used to…
1.Link: http://poj.org/problem?id=1008 http://bailian.openjudge.cn/practice/1008/ 2.content: Maya Calendar Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 66971   Accepted: 20644 Description During his last sabbatical, professor M. A. Ya…
1.链接地址: http://poj.org/problem?id=1094 http://bailian.openjudge.cn/practice/1094 2.题目: Sorting It All Out Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 25547   Accepted: 8861 Description An ascending sorted sequence of distinct values…
1.链接地址: http://bailian.openjudge.cn/practice/1042/ http://poj.org/problem?id=1042 2.题目: Gone Fishing Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 27652   Accepted: 8199 Description John is going on a fishing trip. He has h hours avail…
1.题意翻译:        输入两个整数a,b (0<=a,b<=10),计算a+b的值并输出.       其中还提到输出不能有多余的内容,即不能加多空格符号等内容,不然会报Wrong Answer       不能使用文件,不然会报"Runtime Error"或者"Wrong Answer"2.思路:      这个poj用来解释如何编写代码以及提交的例子题目,我也是用此题建立我解题报告的模板.     思路很简单,就是加起来.3.总结:    …
POJ 排序的思想就是根据选取范围的题目的totalSubmittedNumber和totalAcceptedNumber计算一个avgAcceptRate. 每一道题都有一个value,value = acceptedNumber / avgAcceptRate + submittedNumber. 这里用到avgAcceptedRate的原因是考虑到通过的数量站的权重可能比提交的数量占更大的权重,所以给acceptedNumber乘上了一个因子. 当然计算value还有别的方法,比如POJ上…
以下是poj百道水题,新手可以考虑从这里刷起 搜索1002 Fire Net1004 Anagrams by Stack1005 Jugs1008 Gnome Tetravex1091 Knight Moves1101 Gamblers1204 Additive equations 1221 Risk1230 Legendary Pokemon1249 Pushing Boxes 1364 Machine Schedule1368 BOAT1406 Jungle Roads1411 Annive…
0:Enclosure http://poj.openjudge.cn/challenge3/0/ 查看 提交 统计 提问 总时间限制:  1000ms 内存限制:  131072kB 描述 为了防止爆零而加入了一道热身题.大家轻虐- Picks在参加NOI(网上同步赛)时,看到大家都在疯狂讨论“圈地游戏”,于是Picks对这个游戏很!好!奇!他想自己开发一个Bot来玩. 不过Picks水平不行……写一个复杂的程序对他太难了,于是他简化了一下,使得场上只有他写的Bot,且场地大小无限. 圈地游戏…
地址:http://poj.openjudge.cn/practice/C17K/ 题目: C17K:Lying Island 查看 提交 统计 提问 总时间限制:  2000ms 内存限制:  262144kB 描述 There are N people living on Lying Island. Some of them are good guys, while others are bad. It is known that good guys are always telling t…
题目链接:http://poj.openjudge.cn/practice/C15C/ 题意:n 点 m 边 k 天.每条边在某一天会消失(仅仅那一天消失).问每一天有多少对点可以相互到达. 解法:开始不会做,参考的YYN的题解:http://blog.csdn.net/u013368721/article/details/45725181 学习了这种CDQ加并查集的做法,可以说是非常的巧妙了.复杂度可以保证在:O(KlogklogK)的范围. //CDQ + DSU #include <bit…
题面: 传送门:http://poj.openjudge.cn/practice/1009/ Solution DP+DP 首先,我们可以很轻松地求出所有物品都要的情况下的选择方案数,一个简单的满背包DP就好 即:f[i][j]表示前i个物品装满容量为j的背包的方案数. 转移也很简单 f[i][j]=f[i-1][j]+f[i-1][j-w[i]] (i:1~n,j:1~m) (即选和不选的问题) 初始化 f[i][0]=1 (i:[0~n]) (如果背包容量为0,无论如何都有且只有一种方案将其…
题目链接: http://poj.openjudge.cn/practice/1055/ 题目意思: 给出的树最大节点个数为n的情况下,求树上点深度的期望. 解题思路: 数学期望公式的推导. 自己先画下nodes=1时 p[1]=1 nodes=2时,p[2]=0.5*1+0.5*2=3/2 nodes=3时,p[3]=11/6 nodes=4时,p[4]=50/24 nodes=5时,p[5]=274/120 .......其实p[n]就是调和级数h[n]=1+1/2+1/3+1/4+...1…
MS Recognition 在线提交: hihoCoder 1402 http://hihocoder.com/problemset/problem/1402 类似: OpenJudge - I:PKU Zealots http://poj.openjudge.cn/campus2016/I/ 描述 Given an image containing only two kinds of capital letters, 'M' and 'S', can you tell how many of…
http://poj.openjudge.cn/practice/C18H 题目 算平均数用到公式\[\bar{x}=\frac{x_1+x_2+x_3+\cdots+x_n}{n}\] 但如果用int型计算,那么\(x_1+x_2+x_3+\cdots+x_n\)可能会超过\(2^{31}-1\) 算6个数的平均数可以这么算 Calculate the average of\(x_1,x_2,x_3\)\[\bar{x}_1=\frac{x_1+x_2+x_3}{3}\]Calculate t…