poj 1068 Parencodings

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 16462		Accepted: 9839

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.

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

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

Source

Tehran 2001

      題目大意。記錄括弧序列有兩種方式。p方式。記錄第i個右括弧前有多少個左括弧。w方式。記錄從左至右第i個右括弧和與之配對的左括弧之間有多少個右括弧。包括自己

#include <stdio.h>int main(){    int t,n,i,j,temp,pre,c,cc;//t,n如題意。temp臨時讀資料    int par[50];//通過p序列還原出括弧    int w[25];//用於存w序列    scanf("%d",&t);    while(t--)    {        c=cc=0;//初始化        pre=0;        scanf("%d",&n);        for(i=0; i<n; i++)        {            scanf("%d",&temp);            if(temp!=pre)//判斷左括弧是否增多。若增多則記錄多的左括弧            {                for(j=0; j<temp-pre; j++)                    par[c++]=-1;//-1代表左括弧                pre=temp;//記錄當前左括弧總數            }            par[c++]=1;記錄完左括弧後記錄右括弧        }     //   for(i=0;i<c;i++)     //       printf("%d ",par[i]);//用於檢查記錄是否正確           for(i=0;i<c;i++)           {               if(par[i]==1)//遍曆。先找到右括弧               {                   temp=1;//temp用於計數                   for(j=i-1;j>=0;j--)                   {                       if(par[j]==-1)//若遇到與右括弧配對的左括弧就進行下一次尋找                       {                           par[j]=0;//已配對。刪除左括弧防止幹擾一次尋找                           break;                       }                       if(par[j]==1)//記錄其中包含右括弧的對數                        temp++;                   }                   w[cc++]=temp;               }           }           for(i=0;i<cc;i++)            printf("%d ",w[i]);           printf("\n");    }    return 0;}