递归递推 ZOJ-1633

Big String

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

首先，我们可以通过递推式len[n]=len[n-1]+len[n-2]得出每个长度

又，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的时候输出

感受一个例子：

T.T^__^  ->  T.T^__^T.T  ->  T.T^__^T.TT.T^__^

我们可以发现标红的点都是一个得来的，就可以得到它在前一个字符串的位置：

pos=lower_bound(a,a+89,n)-a;

n-=a[x-1];

^__^

 

#include <cstdio>
#include <string>
#include <algorithm>
using namespace std;

int main()
{
    long long n;
    string c="T.T^__^";
    long long a[105];

    a[0]=4;a[1]=3;
    for(int i=2;i<=88;++i){
        a[i]=a[i-1]+a[i-2];//printf("%lld\n",a[i]);
    }
    while(~scanf("%lld",&n)){
        while(n>7){
            int x=lower_bound(a,a+89,n)-a;
            n-=a[x-1];
        }   //一定相同前缀故而可以直接减！
        printf("%c\n",c[n-1]);
    }

    return 0;
}