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
题意：开始时有两个字符串，A = "^__^" (four characters), B = "T.T" (three characters)，然后重复执行C=BA，A=B,B = C，问第n个字符是什么。
分析：因为最终的字符串是从前往后递推出来的，所以再求解时，我们可以把这个过程逆过去，直到字符串的长度不超过7，输出即可。
#include
#include
using namespace std; typedef long long LL; LL a[95]; //保存字符串的长度 int main() { string C = "T.T^__^"; a[0] = 4; a[1] = 3; int i; for(i = 2; i < 90; i++) a[i] = a[i-1] + a[i-2]; LL n; while(cin >> n) { while(n > 7) { int pos = lower_bound(a, a+89, n) - a; n -= a[pos-1]; } cout << C[n-1] << endl; } return 0; }