Parencodings Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28860   Accepted: 16997 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…

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. 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[] = ;…

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…

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…

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…

Parencodings 题意: 由括号序列S可经P规则和W规则变形为P序列和W序列. p规则是:pi是第i个右括号左边的左括号的数: w规则是:wi是第i右括号与它匹配的左括号之间右括号的数(其中包括它本身). 题解: 这题真的好简单,数据也小,算是一个增加了我信心的题吧. 1是左括号,2是右括号. 代码: #include <vector> #include <cstdio> #include <string> #include <cstdlib> #i…