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
先翻译一下题啊，英语是硬伤啊。本题就是给你n个数。例如6个数，分别为4，5，6，6，6，6.
4表示第一个右括号前面有4个左括号，5表示第二个括号前面有5个左括号。。。
然后题中让输出第一个到最后一个右括号分别包含多少个括号。
如题：
(((()()())))
第一个右括号包含一个括号。第二个右括号包含一个括号。第三个右括号包含一个括号。
第四个右括号包含四个括号。第五个右括号包含五个括号。第六个右括号包含六个括号。
所以输出：1 1 1 4 5 6
此题采用构造法。
我们定义三个数组，const int maxn=50;
int left[maxn];记录左边含有的左括号数
int right[maxn];记录第i个右括号到第i+1个右括号中含有的左括号数
int num[maxn];记录最终要输出的结果。
然后通过right[maxn]数组来得出num[maxn]的值。
代码：
#include<stdio.h>
#include<iostream>
using namespace std;
const int maxn=50;
int main()
{
int left[maxn];
int right[maxn];
int num[maxn];
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
for(int i=1;i<=n;i++)
cin>>left[i];
right[0]=left[1];
for(int i=1;i<n;i++)
{
right[i]=left[i+1]-left[i];
}
int j;
for(int i=1;i<=n;i++)
{
for(j=i-1;j>=0;j--)
{
if(right[j]>0)
{
right[j]--;
break;
}
}
num[i]=i-j;
}
for(int i=1;i<=n;i++)
cout<<num[i]<<" ";
cout<<endl;
}
return 0;
}
巧妙！