本文主要是介绍poj 1068 Parencodings，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
模拟的题型，基本难度不大，关键读懂题意：
对于给出的原括号串，存在两种数字密码串：
1.p序列：当出现匹配括号对时，从该括号对的右括号开始往左数，直到最前面的左括号数，就是pi的值。
2.w序列：当出现匹配括号对时，包含在该括号对中的所有右括号数（包括该括号对）,就是wi的值。
题目的要求：对给出的p数字串，求出对应的s串。
串长限制均为20
提示：在处理括号序列时可以使用一个小技巧，把括号序列转化为01序列，左0右1，处理时比较方便
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:
Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.
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
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
Source
#include<stdio.h>
#include<string.h>
int main()
{int t,m,n,i,j,k;int a[40],b[40],c[40];while(~scanf("%d",&t)){while(t--){memset(b,0,sizeof(b));scanf("%d",&m);a[0]=0;c[0]=0;k=1;for(i=1; i<=m; i++){scanf("%d",&a[i]);c[i]=a[i];for(j=1; j<=a[i]-a[i-1]; j++){b[k++]=0;}b[k++]=1;c[i]=c[i-1]+j;}for(i=1; i<=m; i++){for(j=c[i]-1; j>=0; j--){if(b[j]==0){b[c[i]]++;b[j]=1;break;}elseb[c[i]]++;}b[c[i]]--;}for(i=1; i<=m; i++)printf("%d ",(b[c[i]]+1)/2);printf("\n");}}return 0;
}
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