Parencodings
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 24932 | Accepted: 14695 |
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).
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
解析：模拟。
#include <cstdio> char s[10000000];
int a[25];
int n;
int res[25];
int rid[25]; void solve()
{
int cnt = 0, rcnt = 0;
for(int i = 1; i <= a[0]; ++i)
s[cnt++] = '(';
s[cnt] = ')';
rid[rcnt++] = cnt;
++cnt;
for(int i = 1; i < n; ++i){
int num = a[i]-a[i-1];
for(int j = 1; j <= num; ++j)
s[cnt++] = '(';
s[cnt] = ')';
rid[rcnt++] = cnt;
++cnt;
}
int res_cnt = 0;
for(int i = 0; i < rcnt; ++i){
int l = 0, r = 1;
for(int j = rid[i]-1; ; --j){
if(s[j] == '('){
++l;
if(l == r){
res[res_cnt++] = l;
break;
}
}
else
++r;
}
}
for(int i = 0; i < n-1; ++i)
printf("%d ", res[i]);
printf("%d\n", res[n-1]);
} int main()
{
int t;
scanf("%d", &t);
while(t--){
scanf("%d", &n);
for(int i = 0; i < n; ++i)
scanf("%d", &a[i]);
solve();
}
return 0;
}