[POJ1068]Parencodings 试题描述 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…

Parencodings 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-s…

Parencodings Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 5   Accepted Submission(s) : 5 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Let S = s1 s2 - s2n be a w…

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…

题目链接. 分析: 水题. #include <iostream> #include <cstdio> #include <cstring> using namespace std; ; int P[maxn], W[maxn]; char s[maxn]; int main(){ int T, n, m; scanf("%d", &T); while(T--) { scanf("%d", &n); P[] = ;…

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…

[poj1068] Parencodings Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 26686   Accepted: 15645 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 = p…

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题目传送门 本题知识点:模拟 这是一道恐怖的括号题.题意稍微理解以下还是可以的. 我们针对样例来理解一下 S.P.W 到底是什么意思: S:( ( ( ( ) ( ) ( ) ) ) ) P: \(P_1\) 为 4 :是因为在第一个 ) (数组第4位)前面有4个 ( \(P_2\) 为 5 :是因为在第二个 ) (数组第6位)前面有5个 ( \(P_3\) 为 6 :是因为在第三个 ) (数组第8位)前面有6个 ( (后面 \(P_4\) \(P_5\) \(P_6\) 情况跟 \(P_3\)…