POJ 1068括号模拟

xiaoxiao2021-02-28  18

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 #include<iostream> #include<cstring> using namespace std; int a[100]; char arry[100]; bool book[100]; int main() { int T,n,count,ans; cin>>T; while(T--) { cin>>n; count=0; memset(a,0,sizeof(a)); memset(book,false,sizeof(book)); memset(arry,0,sizeof(arry)); for(int i=1;i<=n;i++) cin>>a[i]; for(int i=0;i<a[1];i++) arry[count++]='('; for(int i=2;i<=n;i++) { if(a[i]==a[i-1]) arry[count++]=')'; else{ arry[count++]=')'; for(int j=a[i-1];j<a[i];j++) arry[count++]='('; } } arry[count++]=')'; //for(int i=0;i<count;i++) cout<<arry[i];// for(int i=0;i<count;i++) { ans=1; if(arry[i]==')') { for(int j=i-1;j>=0;j--) { if(arry[j]==')') ans++; if(arry[j]=='('&&!book[j]) { book[j]=true; break; } } cout<<ans<<" "; } } cout<<endl; } }
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