POJ-1068-Parencodings
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Parencodings
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 24991 Accepted: 14734
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:
Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.
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
#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <stack>#include <queue>using namespace std;int main(){ int n,i,T,x,j,t,k; scanf("%d",&T); while(T--) { scanf("%d",&n); stack<char>S; queue<int>Q; k=0; for(i=0;i<n;i++) { scanf("%d",&x); for(j=k;j<x;j++) { S.push('('); } t=1; k=x; while(!S.empty()) { if(S.top()=='*') { t++; S.pop(); } else break; } S.pop(); Q.push(t); while(t--) { S.push('*'); } } printf("%d",Q.front()); Q.pop(); while(!Q.empty()) { printf(" %d",Q.front()); Q.pop(); } printf("\n"); } return 0;}
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