Big String 二分法的问题

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

题意：很简单就是找给定位置的字符

思路：二分

AC代码：

#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

long long  n;

char c[] = "T.T^__^";

long long a[92];

long long  erfen(){
     long long left = 0,mid;
     long long right = 89;
     while(left < right){
        mid = left + (right - left)/2;
        if(a[mid] >= n){
            right = mid;
        }
        else{
            left = mid + 1;
        }
     }
     return right;//然会大于等于n的数组的下标
}

int main()
{
    a[0] = 4;
    a[1] = 3;//两个顺序不能变啊，wa了n次
    long long i;
    for(i = 2 ; i < 90 ; i++){
        a[i] = a[i-1] + a[i-2];
    }
    while(cin >> n){
        while(n > 7){
           long long mid = erfen();
           n -= a[mid-1];
        }
        cout << c[n-1] << endl;
    }
    return 0;
}

简直就是渣渣，这道题想太长时间，花了太多精力，懂了很多。。。