Parencodings
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions:28556 | Accepted: 16825 |
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
【题意】
给出一个序列P，代表每一个右括号前的左括号数目，要求输出序列W，即每个右括号前能匹配的括号数目。
【解题思路】
用数组b记录每两个右括号之间的左括号数目，然后依次匹配即可。
【代码】
#include<cstdio>
#include<cstring>
using namespace std;
int a[50],b[50],ans[50];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(ans,0,sizeof(ans));
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
b[0]=a[1];
for(int i=1;i<n;i++)
b[i]=a[i+1]-a[i];
for(int i=1;i<=n;i++)
{
int j;
for(j=i-1;j>=0;j--)
{
if(b[j]>0)
{
b[j]--;
break;
}
}
if(j>=0)ans[i]=i-j;
else ans[i]=0;
}
for(int i=1;i<=n;i++)
i==1?printf("%d",ans[i]):printf(" %d",ans[i]);
printf("\n");
}
return 0;
}