POJ-1068-Parencodings-2013-12-12 23:34:16
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Parencodings
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 18245 Accepted: 10981
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 6666W-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
264 5 6 6 6 69 4 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
Tehran 2001
# include<stdio.h># include<string.h>int main(){int s[1000],a[22],t,n,m,i,j,k,num;scanf("%d",&t);while(t--){memset(s,0,sizeof(s));memset(a,0,sizeof(a));scanf("%d",&n);for(i=0;i<n;i++){scanf("%d",&a[i]);if(i==0){for(j=1;j<=a[i];j++){s[j] = 1;}s[j++] = 0;}else{if(a[i]-a[i-1]>0){for(k=0;k<a[i]-a[i-1];k++)s[j++] = 1;s[j++] = 0;}else{s[j++] = 0;}}}for(k=1;k<j;k++){m = k;num = 1;if(s[m]==0){while(1){if(s[m]==0){m--;}else if(s[m]==1){s[m] = 2;if(k!=j-1)printf("%d ",num);elseprintf("%d\n",num);break;}else if(s[m]==2){num++;m--;}}}}}return 0;}
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