f508: 12041 - BFS (Binary Fibonacci String)
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Accepted rate : 9人/12人 ( 75% ) [非即時]

Content

We are familiar with the Fibonacci sequence (1, 1, 2, 3, 5, 8, ...). What if we define a similar sequence for strings? Sounds interesting? Let’s see.
We define the follwing sequence:
BFS(0) = 0 BFS(1) = 1 (here “0” and “1” are strings, not simply the numerical digit, 0 or 1)for all (n > 1) BF S(n) = BF S(n − 2) + BF S(n − 1) (here, ‘+’ denotes the string concatenation operation). (i.e. the n-th string in this sequence is a concatenation of a previous two strings).
So, the first few strings of this sequence are: 0, 1, 01, 101, 01101, and so on.
Your task is to find the N-th string of the sequence and print all of its characters from the i-th to j-th position, inclusive. (All of N, i, j are 0-based indices)

Input

The first line of the input file contains an integer T (T ≤ 100) which denotes the total number of test cases. The description of each test case is given below:
Three integers N, i, j (0 ≤ N, i, j ≤ 2^31 − 1) and (i ≤ j and j − i ≤ 10000). You can assume that,both i and j will be valid indices (i.e. 0 ≤ i, j < length of BFS(N)).

Output

For each test case, print the substring from the i-th to the j-th position of BF S(N) in a single line.

Sample Input #1
3
3 1 2
1 0 0
9 5 12
Sample Output #1
01
1
10101101

Hint ：

2018 12月CPE第七題

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