Parencodings
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 22757 | Accepted: 13337 |
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:
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.
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>
#include <cstdio>
#include <cmath>
#include <string>
#include <stack>
#include <algorithm>
using namespace std; const int Max=1100000; int Arr[110];
bool vis[1100];
int main()
{
int T;
int n;
while(~scanf("%d",&T))
{
while(T--)
{
scanf("%d",&n);
int Max=0;
int sum;
for(int i=0;i<n;i++)
{
scanf("%d",&Arr[i]);
if(Arr[i]>Max)
{
Max=Arr[i];
}
}
Max+=n;
sum=Max;
memset(vis,false,sizeof(vis));
vis[Max]=true;
for(int i=n-2;i>=0;i--)
{ Max-=(Arr[i+1]-Arr[i]+1);
vis[Max]=true;
}
int top=0;
for(int i=1;i<=sum;i++)
{
if(vis[i])
{
int ans=0;
int ant=0;
for(int j=i;j>=1;j--)
{
if(vis[j])
{
ans++;
}
else
{
ans--;
ant++;
}
if(!ans)
{
break;
}
}
Arr[top++]=ant;
}
}
for(int i=0;i<n;i++)
{
if(i)
cout<<" ";
cout<<Arr[i];
}
cout<<endl;
}
}
return 0;
}
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