poj 1068 Parencodings
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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
#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#define INF 0x3f3f3f3f#define LL long longusing namespace std;int p[22] = {0};int Lp[22];int main(){ //freopen("in.txt","r",stdin); int T; scanf("%d",&T); while(T--) { int n; scanf("%d",&n); for(int i=1; i<=n; i++) { scanf("%d",&p[i]); } for(int i=1; i<=n; i++) { Lp[i] = p[i] - p[i-1]; } for(int i=1; i<=n; i++) { int j = i; int ant = 1; while(Lp[j] == 0 && j > 1) { j--; ant++; } Lp[j]--; if(i == n) printf("%d\n",ant); else printf("%d ",ant); } } return 0;}
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