NUC_HomeWork1 -- POJ1068

A - Parencodings

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u

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

今天wj说，这道题模拟就好，用不了多长时间，结果我想了一晚上，用递推的方法做的，不知道他是怎么想的，能A出来，很开心

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 50;

void print(int ans[], int n)///打印函数，其实是在中间查错的时候写的，后来就直接用了
{
     printf("%d", ans[1]);
    for(int i = 2; i <= n; ++i)
    {
        printf(" %d", ans[i]);
    }
    puts("");
}

int main()
{
#ifdef  LOCAL                       ///重定向第一发忘记删了，错了T_T
    freopen("in.txt", "r", stdin);
#endif
    int t, n;
    int arr1[maxn], arr2[maxn], ans[maxn];///我的想法是，预处理arr2数组用以放在此坐标左侧有几个半括号
    bool tag[maxn];                        ///预处理ans数组，每次值变化，都会从“1”开始，然后就处理不是“1”的值
    scanf("%d", &t);                       ///用tag数组标记被处理过的值，
    while(t--)
    {
        memset(tag, false, sizeof(tag));
        scanf("%d", &n);
        for(int i = 1; i <= n; ++i)
            scanf("%d", &arr1[i]);

        ans[1] = 1;
        arr2[1] = arr1[1] - 1;
        for(int i = 2; i <= n; ++i)     ///进行一次预处理,将所有为 “1” 的情形记录
        {
            arr2[i] = arr1[i] - arr1[i-1] - 1;
            arr2[i] = (arr2[i] < 0) ? 0 : arr2[i];
            ans[i] = (arr1[i] > arr1[i-1]) ? 1 : 0;
        }

        int i, k;
        int sum = 0;
        for(i = 2; i <= n; ++i)
        {
            if(ans[i] == 0)
            {
                for(k = i-1; k > 0; --k)              ///往前找无非两种情况可以累加
                {                                      ///没有被标记过的且arr2值为0，和没有被标记过的且arr2值不为0
                    if((tag[k] == false) && (arr2[k] == 0))
                    {
                        sum += ans[k];
                        tag[k] = true;                  ///刚开始也DB了， 用于 “==”，查了半天
                    }
                    else if((tag[k] == false) && arr2[k])
                    {
                        arr2[i] = --arr2[k];
                        arr2[k] = 0;
                        sum += ans[k] + 1;              ///找到arr2值不为‘0’，就arr2值转移到 i 身上。到这步就到底了
                        tag[k] = true;
                        //printf("%d
", arr2[i]);
                        break;
                    }
                }
                ans[i] = sum;
                sum = 0;
            }
        }

       print(ans, n);
    }
    return 0;
}