Parencodings Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 22849 Accepted: 13394 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…
POJ 3252:Round Numbers Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 10099 Accepted: 3669 Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Sciss…
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…
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…
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…
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…
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…
Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 22764 Accepted: 13344 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…
http://poj.org/problem?id=3580 题意:有6种操作,其中有两种之前没做过,就是Revolve操作和Min操作.Revolve一开始想着一个一个删一个一个插,觉得太暴力了,后来发现可以把要放到前面的一段切开,丢到前面去,就和上一题的Cut是一样的了.还有Min操作,一开始特别ZZ地想着只要找keytree的最左边就好了,然后发现并不是那样的,要维护一个 mi 值,一开始两个节点设成 INF,然后 pushup 的时候先把 val 赋给 mi,然后再和左右儿子对比.WA了…
http://poj.org/problem?id=2763 题意:给出 n 个点, n-1 条带权边, 询问是询问 s 到 v 的权值, 修改是修改存储时候的第 i 条边的权值. 思路:树链剖分之修改边权.边权的修改, 与点权修改不同的地方在于, 线段树中存的点是边,其中每条边边是以 儿子 的时间戳来记录的.例如: u -> v , dep[u] < dep[v], 说明 u 是 v 的父亲,所以这条边在线段树中的编号就是以 tid[v]. #include <cstdio> #…
http://poj.org/problem?id=3349 Snowflake Snow Snowflakes Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 37609 Accepted: 9878 Description You may have heard that no two snowflakes are alike. Your task is to write a program to determine…
http://poj.org/problem?id=1260 Pearls Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 8474 Accepted: 4236 Description In Pearlania everybody is fond of pearls. One company, called The Royal Pearl, produces a lot of jewelry with pearls…
http://poj.org/problem?id=3903 Stock Exchange Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 5983 Accepted: 2096 Description The world financial crisis is quite a subject. Some people are more relaxed while others are quite anxious. J…
http://poj.org/problem?id=3267 The Cow Lexicon Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 9380 Accepted: 4469 Description Few know that the cows have their own dictionary with W (1 ≤ W ≤ 600) words, each containing no more 25 of t…
http://poj.org/problem?id=3026 Borg Maze Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 12086 Accepted: 3953 Description The Borg is an immensely powerful race of enhanced humanoids from the delta quadrant of the galaxy. The Borg coll…
http://poj.org/problem?id=2192 Zipper Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 17585 Accepted: 6253 Description Given three strings, you are to determine whether the third string can be formed by combining the characters in the…