【递推】【巧妙模拟】【思维】ZOJ 1633 Big String 【拼接两个字符串】

【递推】【巧妙模拟】【思维】ZOJ 1633 Big String 【拼接两个字符串】

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：
A = "^__^"
B = "T.T"
不断执行
C = BA
B = C
A = B
将AB串拼接起来，构成一个无限长的字符串T.T^__^T.TT.T^__^......
问第n个字符是什么

思路：
因为这个字符串是由两个基本串构成的，因此只需要用基本串的长度3 ,4来模拟拼接过程，然后判断第n个字符在基本串的哪里即可。

• f[i]表示第i次拼接后字符串的长度
• 模拟拼接过程 即递归过程 f[i] = f[i - 1] + f[i - 2]
• 判断第n个字符在基本串的哪里：找到第一个小于n的字符串长度（使用low_bound），这个字符串就是拼接在第n位前面的子串，$\color {green}{逆拼接过程}$，将这个子串去掉（也就是n -= f[i]），一直重复该过程，直到n < 7，即可以在标准串T.T^__^中找到他的位置。

AC代码：

#include <iostream>
#include <stdio.h>
#include <algorithm>

using namespace std;

const int maxn = 100;
long long f[maxn];

int main()
{
    string c = "T.T^__^";
    f[0] = 4;
    f[1] = 3;
    for(int i = 2; i < 90; i++)
        f[i] = f[i - 1] + f[i - 2];
    long long n;
    while(~scanf("%lld", &n))
    {
        while(n > 7)
        {
            int p = lower_bound(f, f + 90, n) - f;
            n -= f[p - 1];
        }
        printf("%c\n", c[n - 1]);
    }
    return 0;
}