POJ 1068 Parencodings 模拟 难度:0】的更多相关文章

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: 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…
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…
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=2632 #include<cstdio> #include <cstring> #include <algorithm> using namespace std; int A,B,n,m; int robot[101][3]; char rbuff[10]; int dir[255]; const int dx[4]={0,1,0,-1}; const int dy[4]={1,0,-1,0}; int action…
#define ONLINE_JUDGE #include<cstdio> #include <cstring> #include <algorithm> using namespace std; int A,B,sx,sy; char maz[101][101]; int vis[101][101]; const int dx[4]={0,1,0,-1}; const int dy[4]={-1,0,1,0}; int dir(char ch){ if(ch=='N'…
Emag eht htiw Em Pleh Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2806   Accepted: 1865 Description This problem is a reverse case of the problem 2996. You are given the output of the problem H and your task is to find the correspond…