Parencodings（poj1068简单模拟）

/*http://acm.hust.edu.cn/vjudge/contest/view.action?cid=27454#problem/B
B - Parencodings
Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u
Submit
 
Status
 
Practice
 
POJ 1068
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
解析：
题意：
给出一组数，第i个数表示第i个右括号前面有几个左括号
要求输出的一组数据，表示每个右括号与它前面的第几个左括号匹配
思路：
1.建模：根据所给出数据构造一组数据用1表示右括号，0表示左括号.s[i]表示第i个值是哪种括号
2,查找：从每个右括号向前找出未被匹配的左括号，匹配完毕要做标记
Memory Time Language Length Submit Time
164 KB 0 ms C++ 641 B 2013-07-29 09:25:05
*/

#include<stdio.h>#include<string.h>#include <iostream>using namespace std;int s[100],q[25],w[25];int main(){  int T,n,q1,i,j,k;scanf("%d",&T);while(T--){   memset(s,0,sizeof(s));scanf("%d",&n);scanf("%d",&q[1]);q1=q[1];s[q1+1]=1;k=q1+1;for(i=2;i<=n;i++){scanf("%d",&q[i]);k+=q[i]-q1+1;s[k]=1;q1=q[i];}memset(w,0,sizeof(w));for(i=1;i<=n;i++) {k=0; for(j=q[i]+i-1;j>=1;j--) { if(s[j]==0) {k++;  s[j]=-1; break; } if(s[j]==-1) k++; } w[i]=k; } for(i=1;i<=n;i++) {printf("%d",w[i]); if(i!=n)   printf(" ");   else   printf("\n"); }}return 0;}