POJ-1068解题报告

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Language:
Parencodings
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 13224 Accepted: 7865
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
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#include <stdio.h>int main(){    int n;    int m, i, p, c;    int a[20];    int arr[20];    scanf("%d", &n);    while (n-- > 0) {        p = 0;        scanf("%d", &m);        for (i = 0; i < m; i++) {            scanf("%d", &a[i]);            arr[i] = !i ? a[i] : a[i] - a[i - 1];            p = arr[i] ? i : p;            printf("%d", 1 + i - p);            arr[p]--;            while (!arr[p]) {                p--;            }            if (i < m - 1) {                printf(" ");            } else {                printf("\n");            }        }    }    return 0;}

这道题还是比较简单的，可以在输入过程就把输出给确定了。每输入一个数，此位置之前有几个左括号，一个p标识符用于标识离当前位置最近的（包括当前位置）左括号数大于0的右括号位置，输出的值是当前位置与p位置差加1。具体原理就不多说了，看源码应该能比较容易理解。