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 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
这个题理解了题意就不难了
输入2 6 不解释
4 5 6 6 6 6
可以这样理解
(((()()())))
第一个右括号左边有4个左括号
第二个右括号左边有5个左括号
以此类推;
所以根据所给出的数字串;很容易还原括号;
输出时:(((()()())))
一个完整的()输出1;共3个
(()()())这个输出4一个()有3个(),3+1;
一句话就是
比如(((()()()))),
输入:每个右括号之前的左括号数序列为P=4 5 6 6 6 6,;
输出:每个右括号所在的括号内包含的括号数为W=1 1 1 4 5 6.
可以利用 栈和队列
1 #include<iostream> 2 #include<queue> 3 #include<stack> 4 #include<cstdio> 5 using namespace std; 6 int main() 7 { 8 queue<char>p,q; 9 stack<char>Q; 10 int a,b,c,m,n; 11 cin>>n; 12 while(n--) 13 { 14 cin>>m; 15 b=0; 16 while(m--) 17 { 18 cin>>a; 19 c=a-b;//制定压入的左括号的数目 20 b=a; 21 while(c--) 22 p.push('('); 23 p.push(')');//每次压入左括号之后压入一个右括号 24 } 25 while(!p.empty())//将队列中的括号一个个取出 26 { 27 if(p.front()=='(') 28 Q.push(p.front());//如果是左括号直接压入栈中 29 30 else if(p.front()==')')//如果是右括号,则进行判断 31 { 32 if(Q.top()=='(')//如果栈顶元素是左括号 33 { 34 Q.pop();//出栈 35 Q.push(1);//压入 1 36 q.push(1);//这是为了把数字储存下来,方便以后输出答案 37 } 38 else//如果栈顶元素不是左括号 39 { 40 int sum=1; 41 while(Q.top()!='(')//持续出栈 直到左括号 42 { 43 sum+=Q.top();//统计一共包含多少左括号 44 Q.pop(); 45 } 46 Q.pop();//左括号出栈 47 q.push(sum);//这是为了把数字储存下来,方便以后输出答案 48 Q.push(sum);//压入括号个数 49 } 50 } 51 p.pop(); 52 } 53 printf("%d",q.front()); 54 q.pop(); 55 while(!q.empty()) 56 { 57 printf(" %d",q.front()); 58 q.pop(); 59 } 60 printf(" "); 61 } 62 return 0; 63 }