(模拟) poj 1068

时间：2015-05-03 18:51:45 阅读：93 评论：0 收藏：0 [点我收藏+]

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Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 20999		Accepted: 12585

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
int n,a[21],sum[21];
int main()
{
    int tt;
    scanf("%d",&tt);
    while(tt--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=1;i<=n;i++)
            sum[i]=a[i]-a[i-1];
        for(int i=1;i<=n;i++)
        {
            int j=i;
            while(j>1&&!sum[j])
                j--;
            sum[j]--;
            printf("%d%c",i-j+1,i==n?‘\n‘:‘ ‘);
        }
    }
    return 0;
}

(模拟) poj 1068

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原文地址：http://www.cnblogs.com/water-full/p/4474309.html

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