HDU-5351

MZL's Border

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 671    Accepted Submission(s): 209

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

 

Source

2015 Multi-University Training Contest 5

/**
     题意：给一个A串，一个B串，然后str[i] = str[i-1]+str[i-2];
            求解对于第n个串的第前m位最长的LBorder，LBorder 是指s[1:i]=s[n−i+1:n],
    做法：画几个串可以找到规律，比如第九个串
        a    b a   a b a    b a a b a   a b a b a a b  a    b a a  b  a a  b  a   b a  a  b  a
        [0] [0 1] [1 2 3]  [2 3 4 5 6] [4 5 6 7 8 9 10 11] [7 8 9 10 11 12 13 14 15 16 17 18 19]
        Java 
 **/
import java.util.*;
import java.math.*;

public class Main 
{
    public static void main(String args[]) 
    {
        BigInteger AA[] = new BigInteger [1000+5];
        BigInteger A[] = new BigInteger [1000+5];
        A[1] = BigInteger.ONE;
        A[2] = BigInteger.valueOf(2);
        BigInteger MOD = BigInteger.valueOf(258280327);
        BigInteger mm = BigInteger.valueOf(1);
        for (int i = 3; i <= 1001; i++) {
            A[i] = A[i-1].add(A[i-2]);
        }
        for(int i=2;i<=1001;i++)
        {
            A[i] = A[i].add(A[i-1]);
        }
        AA[1] = BigInteger.ZERO;
        AA[2] = BigInteger.ZERO;
        for(int i=3;i<=1001;i++)
        {
            AA[i] = AA[i-1].add(AA[i-2]).add(mm);
        }
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        while (t-- > 0) 
        {
            int n = in.nextInt();
            BigInteger m = in.nextBigInteger();
            if (m.equals(BigInteger.valueOf(1)) || m.equals(BigInteger.valueOf(2)))
            {
                System.out.println(0);
                continue;
            }
            boolean ok = false;
            int tt = 0;
            for (int i = 1; i <= 1001; i++) 
            {
              int yy = A[i].compareTo(m);
              if(yy == 1) break;
              tt = i;
            }
            if(A[tt].compareTo(m) == 0)
            {
                 BigInteger temp = BigInteger.ZERO;
                 BigInteger res = A[tt].subtract(A[tt-1]);
                 BigInteger tmp  = AA[tt].add(res) ;
                 tmp = tmp.subtract(mm);
                 System.out.println(tmp.mod(MOD));
            }
            else 
            {
                BigInteger temp = BigInteger.ZERO;
                BigInteger res = m.subtract(A[tt]);
                BigInteger tmp  = AA[tt+1].add(res);
                tmp = tmp.subtract(mm);
                System.out.println(tmp.mod(MOD));
            }
          }   
    }
}