Parencodings - POJ 1068 水题

Parencodings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 19336 Accepted: 11665

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 6666W-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.

Outputb the output line should contain n integers describing the W-sequence of the string corresponding to its given P-sequence.

Sample Input

264 5 6 6 6 69 4 6 6 6 6 8 9 9 9

Sample Output

1 1 1 4 5 61 1 2 4 5 1 1 3 9

题意：第一种是表示方法是第几个）左边有多少（，第二种表示方法是第几个）和与其对应的（中间的（的数量。

思路：先还原数据，在求括号的对应。

AC代码如下：

#include<cstdio>#include<cstring>using namespace std;int num[500],sum[500],f[500],p[500],q[500];int main(){ int t,n,i,j,k,pos,pos2;  scanf("%d",&t);  while(t--)  { scanf("%d",&n);    pos=0;    pos2=0;    for(i=1;i<=n;i++)    { scanf("%d",&num[i]);      for(j=num[i-1];j<num[i];j++)      { f[++pos]=1;        p[++pos2]=pos;      }      f[++pos]=2;      q[pos]=p[pos2--];    }    for(i=1;i<=2*n;i++)     if(f[i]==1)      sum[i]=sum[i-1]+1;     else      sum[i]=sum[i-1];     for(i=1;i<=2*n-1;i++)      if(f[i]==2)       printf("%d ",sum[i]-sum[q[i]-1]);    printf("%d\n",sum[2*n]-sum[q[2*n]-1]);  }}