Parencodings
Time Limit:1000MS Memory Limit:65536K
Total Submit:176 Accepted:113
Description
Let S = s1 s2 … s2n be a well-formed string of parentheses. S can be encoded in two different ways:
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).
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 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
264 5 6 6 6 694 6 6 6 6 8 9 9 9
Sample Output
1 1 1 4 5 61 1 2 4 5 1 1 3 9
Source
Asia 2001, Tehran (Iran)
题目其实很简单,只是i前面有p[i]个“)”,求i前面有几个“()”
代码:
#include<stdio.h>main(){ int t,n,i,j,k,s[40],f,c,count; while(scanf("%d",&t)!=EOF) { for(i=1;i<=t;i++) { scanf("%d",&n); int num[n+1]; f=1; num[0]=0; for(j=1;j<=n;j++) { scanf("%d",&num[j]); for(k=1;k<=num[j]-num[j-1];k++) { s[f]=0; f++; } s[f]=1; f++; } f=f-1; for(j=1;j<=n;j++) { for(k=1;k<=f;k++) { if(s[k]==1) { s[k]=2; for(c=k-1;c>=1;c--) { if(s[c]==0) { s[c]=2; count=(k-c+1)/2; printf("%d ",count); break; } } } } } printf("\n"); } }}
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