HDU 5351——MZL's Border——————【高精度+找规律】

MZL's Border

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 944    Accepted Submission(s): 306

Problem Description

As is known to all, MZL is an extraordinarily lovely girl. One day, MZL was playing with her favorite data structure, strings.

MZL is really like Fibonacci Sequence, so she defines Fibonacci Strings in the similar way. The definition of Fibonacci Strings is given below.
  
  1) fib1=b
  
  2) fib2=a
  
  3) fibi=fibi−1fibi−2, i>2
  
For instance, fib3=ab, fib4=aba, fib5=abaab.

Assume that a string s whose length is n is s1s2s3...sn. Then sisi+1si+2si+3...sj is called as a substring of s, which is written as s[i:j].

Assume that i<n. If s[1:i]=s[n−i+1:n], then s[1:i] is called as a Border of s. In Borders of s, the longest Border is called as s' LBorder. Moreover, s[1:i]'s LBorder is called as LBorderi.

Now you are given 2 numbers n and m. MZL wonders what LBorderm of fibn is. For the number can be very big, you should just output the number modulo 258280327(=2×317+1).

Note that 1≤T≤100, 1≤n≤103, 1≤m≤|fibn|.

 

Input

The first line of the input is a number T, which means the number of test cases.

Then for the following T lines, each has two positive integers n and m, whose meanings are described in the description.

 

Output

The output consists of T lines. Each has one number, meaning fibn's LBorderm modulo 258280327(=2×317+1).

 

Sample Input

2

4 3

5 5

 

Sample Output

1

2

 

题目大意：这个题目定义了一些新概念，而且比较绕，所以题意不是太好理解。给T组测试数据，每组有n,m。表示从fibn这个字符串中截取子串s[1:m]。然后在这个字串中找到最大的i满足s[1:i]=s[m-i+1:m]，即公共前后缀。当时理解成了跟n有关系，即s[1:i]=s[n-i+1:n]，主要受那个题的描述影响了。

 

解题思路：其实这个题目计算只跟m有关系，因为n只是进行限制了m的大小，跟n没多大关系。我们列出几项会发现，m跟LBorder_m的差值是成斐波那契数列的。

 

import java.io.*;
import java.util.*;
import java.math.*;
public class Main{
	public static void main(String [] args){
		BigInteger one=BigInteger.ONE, zero = BigInteger.ZERO ;
		BigInteger fib[] = new BigInteger [1200];
		fib[1]=one;
		fib[2]=one;
		for(int i=3;i<=1005;i++){
			fib[i]=fib[i-1].add( fib[i-2] );
		}
		Scanner cin = new Scanner (System.in);
		int t=cin.nextInt();
		while(t>0){
			t--;
			int n= cin.nextInt();
			BigInteger m =  cin.nextBigInteger();
			for(int i=1;i<=1005;i++){
				if(fib[i].compareTo(m.add( one ) )==1){
					System.out.println( m.subtract(fib[i-2]).mod(BigInteger.valueOf(258280327 )));
					break;
				}
			}
		}
	}
}