zoj1633 Big String dfs

Big String
Time Limit: 2 Seconds      Memory Limit: 65536 KB
We will construct an infinitely long string from two short strings: A = "^__^" (four characters), and B = "T.T" (three characters). Repeat the following steps:
Concatenate A after B to obtain a new string C. For example, if A = "^__^" and B = "T.T", then C = BA = "T.T^__^".
Let A = B, B = C -- as the example above A = "T.T", B = "T.T^__^".
Your task is to find out the n-th character of this infinite string.



Input
The input contains multiple test cases, each contains only one integer N (1 <= N <= 2^63 - 1). Proceed to the end of file.


Output
For each test case, print one character on each line, which is the N-th (index begins with 1) character of this infinite string.


Sample Input

1
2
4
8

Sample Output

T
.
^
T

import java.io.BufferedInputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.Scanner;

public class Main {

public static void main(String[] args) {
new Task().solve();
}
}

class Task {
Scanner in = new Scanner(new BufferedInputStream(System.in)) ;
PrintWriter out = new PrintWriter(System.out);

final int N = 90 ;
BigInteger[] fibo = new BigInteger[N+1] ;
char[] res = {' ' , 'T' , '.' , 'T' , '^' , '_' , '_' , '^'} ;

void dfs(BigInteger n){
if(n.compareTo(BigInteger.valueOf(7)) <= 0){
out.println(res[n.intValue()]) ;
return ;
}
int i = 1 ;
for(; i <= N ; i++){
if(fibo[i].compareTo(n) >= 0){
break ;
}
}
dfs(n.subtract(fibo[i-1])) ;
}

void solve() {
fibo[1] = BigInteger.valueOf(4) ;
fibo[2] = BigInteger.valueOf(3) ;
for(int i = 3 ; i <= N ; i++){
fibo[i] = fibo[i-1].add(fibo[i-2]) ;
}
while(in.hasNext()){
dfs(in.nextBigInteger()) ;//out.flush();
}
out.flush();
}
}