Parencodings
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 28860 |
|
Accepted: 16997 |
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
从别人的博客上学的：
https://www.cnblogs.com/--ZHIYUAN/p/5910981.html
#include<iostream>
#include<string>
#include<cstdio>
using namespace std;
int main(){
int n,T;
scanf("%d",&T);
while(T--){
int q1,q2 = 0;
string s;
scanf("%d",&n);
for(int i = 0; i < n; i++){
scanf("%d",&q1);
q2 = q1 - q2;
while(q2--)
s += "(";
s += ")";
q2 = q1;
}
int num = s.length();
int cnt = 0;
int sum;
for(int i = 0; i < num; i++){
if(s[i] == ‘)‘){
cnt++;
int flag = 1;
sum = 0;
for(int j = i - 1; j >= 0; j--){
if(flag == 0)
break;
if(s[j] == ‘)‘){
flag += 1;
}
else if(s[j] == ‘(‘){
flag -= 1;
sum++;
}
}
if(cnt != n)
printf("%d ",sum);
else
printf("%d\n",sum);
}
}
}
return 0;
}
View Code
PO1068 Parencodings 模拟题
原文：https://www.cnblogs.com/Weixu-Liu/p/10422650.html
