poj1068

Parencodings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 24641 Accepted: 14491

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 By an integer sequence W = w1 w2...wn where for each right parenthesis, say a in S, we associate an integer which is the number of right parentheses counting from the matched left parenthesis of a up to a. (W-sequence).

Following is an example of the above encodings:

S(((()()())))P-sequence    4 5 6666W-sequence    1 1 1456

Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.

Input

The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case is an integer n (1 <= n <= 20), and the second line is the P-sequence of a well-formed string. It contains n positive integers, separated with blanks, representing the P-sequence.

Output

The output file consists of exactly t lines corresponding to test cases. For each test case, the output line should contain n integers describing the W-sequence of the string corresponding to its given P-sequence.

Sample Input

264 5 6 6 6 69 4 6 6 6 6 8 9 9 9

Sample Output

1 1 1 4 5 61 1 2 4 5 1 1 3 9

Source

Tehran 2001

个人感觉编程能力实在是弱。一题简单模拟都卡了好久，还要抄网上的程序。

题意如下，给定一个完整的括号序列（每个左括号都有一个右括号和它对应），输入n行，输入这个序列第i个右括号前总共有几个左括号，求每个对应左右括号里面的左右括号数（包括它本身）。

思路如下：既然输入的是每个右括号之前所有的左括号数，那么久每读入一个数据，记录下A(i)-A(i-1)个左括号，再记录一个右括号。输出时，用递归法，若该括号是左括号，证明其可以再次向右递归，沿途向右递归记录下内部的括号总数，知道遇到了它的“另一半”，则结束递归，记录当前值。

15773121 ksq20131068Accepted696K0MSG++871B2016-07-21 13:18:24

#include<vector>#include<cstdio>#include<iostream>using namespace std;int n,cnt,w[10000];vector<bool>bin;void in(){    pair<int,int>p;    p.first=p.second=0;    for(int i=0;i<n;i++){        scanf("%d",&p.second);        for(int j=1;j<=p.second-p.first;j++)bin.push_back(1);        bin.push_back(0);        p.first=p.second;    }}int solve(int &l){    int res=1;    for(;;){        if(bin[l]){            l++;            res+=solve(l);        }        else{            l++;            w[cnt++]=res;            return res;        }    }}void out(){    printf("%d",w[0]);    for(int i=1;i<n;i++)        printf(" %d",w[i]);    putchar('\n');}int main(){    int T,l;    scanf("%d",&T);    for(;T;T--){        l=cnt=0;        bin.clear();        scanf("%d",&n);        in();        solve(l);        out();    }    return 0;}