[ACM_模拟] POJ1068 Parencodings （两种括号编码转化 规律 模拟）

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

Source

Tehran 2001

 

题目大意：一组标准的括号（就是每一个左括号的都有且仅有一个右边的对应，符合常理），其编码方式有两种：

                  P:p[i]表示第i个右括号左边有的左括号数量

                  W:w[i]表示从与第i个右括号对应的左括号开始至第i个右括号共有右括号的数量

              现在给出P串输出W串

解题思路：用b[]统计第i个右括号和第i-1个右括号之间有多少个左括号，然后模拟流程（对于每个右括号向前找

最近左括号，然后在找到的左括号对应区间的b[]减1，依次寻找....

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
int main(){
    int n,a[25],b[25]; 
    int t;cin>>t;
    while(t--){
        cin>>n;              //输入
        for(int i=0;i<n;i++)
            cin>>a[i];
        b[0]=a[0];           //求b[]
        for(int i=1;i<n;i++)
            b[i]=a[i]-a[i-1];
        cout<<1;             //求解并输出
        for(int i=1;i<n;i++){
            for(int j=i;j>=0;j--){
                if(b[j]!=0){
                    cout<<' '<<i-j+1;
                    b[j]--;
                    break;
                }
            }
        }cout<<'\n';
    }return 0;
}