[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…

题目链接: https://vjudge.net/problem/POJ-1068 题目大意: 给出一种括号序列的表示形式名叫P序列,规则是统计出每个右括号之前的左括号个数作为序列每项的值.然后要求你根据这个求括号列的W序列值,W序列的规则是统计每一个右括号和与其匹配的左括号之间所有匹配后的括号个数. 思路: 先模拟出括号序列,然后根据每个右括号向前入栈,记录中间的括号数目 #include<iostream> #include<algorithm> #include<cst…

下面的代码是北京大学Online Judge网站上1068题(网址:http://poj.org/problem?id=1068)的所写的代码. 该题的难点在于实现括号匹配,我在代码中采取用-1和1分别代表左右括号,使得括号匹配时各位数相加为0,不匹配时则不为0的方法来判断是否完成括号匹配,代码列表如下: 性能:Memory:704K,Time:0MS #include "iostream" using namespace std; int main() { const int LEF…

Parencodings Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 18785   Accepted: 11320 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…

A - Parencodings Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u Submit Status 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…

Parencodings Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 25010   Accepted: 14745 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 1068,题目链接http://poj.org/problem?id=1068 题意: 对于给出给出的原括号串S,对应两种数字密码串P.W: S         (((()()()))) P-sequence      4 5 6666   (Pi表示第i个右括号前面有多少个左括号) W-sequence     1 1 1456    (Wi表示第i个右括号对应它前面的第几个左括号) 要求给出P串,求W. 思路: 1. 模拟类题型.将输入的P串先装换为S串,再由S串得到W串. 2. 左…

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…

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[] = ;…