Parencodings
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 24202 | Accepted: 14201 |
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:
Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.
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
http://blog.csdn.net/xinghongduo/article/details/6174671
http://blog.chinaunix.net/uid-22609852-id-3506161.html
ac代码:
#include <stdio.h> #include <stdlib.h> char c_kuohao[10000] = {0}; //生成空格匹配的字符串 void genkuohao(char* c_kuohao,int* array,int arraylength ) { int cur_index = 0; for (int i = 0; i< arraylength-1;i++) { int j ; for ( j = 0;j <*(array+i+1)- *(array +i);j++) { c_kuohao[cur_index + j] = '('; } c_kuohao[cur_index + j ] = ')'; cur_index = cur_index + j +1; } } //从括号字符串中,获取int 数组 //找到一个右括号,把匹配最近的左括号设置为字符1,并生成对应的rarray数组 void getWarray(char* c_kuohao,int* rarray,int arraylength) { int index = 0; int i = 0; while(c_kuohao[i]!=0) { if (c_kuohao[i] ==')') { int j = i-1; while(c_kuohao[j]!='(') { j--; if (c_kuohao[j] == '1') { *(rarray + index) += 1; } } *(rarray + index) += 1; c_kuohao[j] ='1' ; index++; i++; } else { i++; } } } void main() { //freopen("sample.in", "r", stdin); //freopen("sample.out", "w", stdout); /* 同控制台输入输出 */ int mainIndex = 0; scanf("%d",&mainIndex); for (int i = 0; i < mainIndex;i++) { int N = 0; scanf("%d",&N); // 下面申请内存时候要用sizeof不然free时候会算错导致堆出错 int *array = (int*)malloc(sizeof(int)*(N +1)); int *rarray = (int*)malloc(sizeof(int)*N); //给数组第一个位置放个0 *(array+0) = 0; for (int j = 1;j<=N;j++) { scanf("%d",array+j); *(rarray + j-1) =0; } for (int k = 0;k<10000;k++) { c_kuohao[k] = 0; } genkuohao(c_kuohao,array,N+1); getWarray(c_kuohao,rarray,N); for (int z = 0;z<N;z++) { printf("%d ",*(rarray + z)); } printf(" "); free(array); free(rarray); } }
再分享一个非常短的代码:
http://blog.csdn.net/qingniaofy/article/details/7701626