ZOJ - 1633 Big String

Description

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

1248

Sample Output

T.^T

分析：本题的规律很明显，第三个串由前两个串拼接，这和非波拉其数列的定义一致，所以我们可以先求出串长的非波拉其数列，存放在数组中，然后求解每组数据；由于当前串是前两个串拼接而来，对于给定的n，可以找到大于n的最小的串长，那么n一定在前两个串中，假设紧邻的三个串分别是A，B，C，当前n在C中，由于n大于B，所以n一定在A中，由题意，A在B后，所以n-len(B)即在A中的位置，如此循环直到n小于7.

#include <cstdio>using namespace std;const int maxn = 1000000;const char* str = "T.T^__^";typedef long long LL;LL fibo[maxn];int main(){LL n;fibo[0] = 4; fibo[1] = 3;for(int i=2; i<maxn; i++) {fibo[i] = fibo[i-1] + fibo[i-2];}while(~scanf("%llu",&n)) {while(n>7) {int i = 0;while(i<maxn && fibo[i]<n) i++;n -= fibo[i-1];}printf("%c\n", str[n-1]);}    return 0;}