poj 1068 Parencodings(模拟)

poj 1068 Parencodings(模拟)

题目：​​class="data-table" data-id="t7a7e9d1-DjAAm2l3" data-transient-attributes="class" data-width="1162px" style="width: 100%; outline: none; border-collapse: collapse;">

Time Limit: 1000MS

Memory Limit: 10000K

Total Submissions: 23127

Accepted: 13551

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 6666 W-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

2

6

4 5 6 6 6 6

9

4 6 6 6 6 8 9 9 9

Sample Output

1 1 1 4 5 6

1 1 2 4 5 1 1 3 9

英语不好，理解题意困难||-_-

P sequence：第i个右括弧左边有ai个左括弧。

W sequence：第i个右括弧的匹配括弧对数ai=第i个右括弧与它相应的左括弧包含的已匹配的括号对数加上1（新匹配的）

模拟：

#include #include #include using namespace std;const int maxn=5005;char str[maxn];int cnt,ans[maxn],top;int main(){ //freopen("cin.txt","r",stdin); int t,n,old,a; cin>>t; while(t--){ scanf("%d",&n); memset(str,0,sizeof(str)); top=cnt=old=0; for(int i=0;i=0;j--){ if(str[j]==1)res++; else if(str[j]=='('){ res++; str[j]=1; break; } } ans[top++]=res; } } for(int i=0;i

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