POJ1068

日常刷水题

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 24313		Accepted: 14277

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

#include <iostream>

using namespace std;

int a[10000],b[10000],l;

void huanyuan(){

  int i,j,k=0;

  for(i=0;i<a[0];i++){

      b[k]=1;
 k++;

  }

  b[k]=2;

  k++;

  for(i=1;i<l;i++){
 if(a[i]==a[i-1]){

         b[k]=2;
k++;
 }
 else{
 for(j=0;j<a[i]-a[i-1];j++){

           b[k]=1;
   k++;
 }
 b[k]=2;
  k++;
 }

  }

  /*for(i=0;i<l*2;i++){

   if(b[i]==1)
  cout<<"(";

     else
cout<<")";

 

  }*/

}

void jisuan(){

int i,j,k,m;

for(i=0;i<l*2;i++){
if(b[i]==2){
k=1;
m=0;
for(j=i-1;j>=0;j--){
if(b[j]==1){

              k--;
 m++;
}
else{

            k++;
}

          if(k==0)
 break;
}
cout<<m<<" ";
}

}

cout<<endl;

}

int main(){

int n,i,k=0;

cin>>n;

while(n>0){

n--;

cin>>l;

for(i=0;i<l;i++)

 cin>>a[i];

huanyuan();

jisuan();

}

return 0;

}