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[100];
string base="T.T^__^";
void init(){
   x[2]=10;
   x[1]=7;
   for(int i=3;i<100;i++)
   {
       x[i]=x[i-1]+x[i-2];
   }
}
int main()
{
   long long n;
   init();
   while(cin>>n){
       while(n>7){
        int i=0;
        while(i<100&&x[i]<n)
            i++;
        n-=x[i-1];
       }
       cout<<base[n-1]<<endl;
   }
   return 0;
}