ZOJ-Big string（服气思维）

个人心得：我在分治上看到的，但是感觉跟分治没关系，一眼想到斐波那契数可以找到此时n的字符串，但是无法精确到字母，题解的思路

真是令人佩服，以BA为基准，然后只要此时的长度大于7那么必然可以减去最大的斐波那契数然后转换为基准，此时直接输出就好了。服气服气

题目：

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

 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<iomanip>
 #include<algorithm>
 using namespace std;
 #define maxi 0x7FFFFFFFFFFFFFFFLL
 long long x[];
 string base="T.T^__^";
 void init(){
    x[]=;
    x[]=;
    for(int i=;i<;i++)
    {
        x[i]=x[i-]+x[i-];
    }
 }
 int main()
 {
    long long n;
    init();
    while(cin>>n){
        while(n>){
         int i=;
         while(i<&&x[i]<n)
             i++;
         n-=x[i-];
        }
        cout<<base[n-]<<endl;
    }
    return ;
 }