HDU 5351 MZL's Border

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
  
  

  
  
   找找規律，然後模拟一下,注意有大數
  
  
          
import java.io.*;
import java.math.BigInteger;
import java.util.*;

public class Main
{
    public static void main(String args[])
    {
        Scanner cin = new Scanner(System.in);    
        BigInteger f[][] = new BigInteger[1005][3];
        int i;
        f[0][1]=BigInteger.ZERO;
        f[0][2]=BigInteger.ZERO;
        f[1][1]=BigInteger.ZERO;
        f[1][2]=BigInteger.ZERO;
        f[0][0]=BigInteger.ONE;
        f[1][0]=BigInteger.ONE;
        for (i=2;i<=1000;i++)
        {
            f[i][0]=f[i-1][0].add(f[i-2][0]);
            f[i][1]=f[i-1][2].subtract(f[i-1][1]);
            f[i][2]=f[i][1].add(f[i][0]).subtract(BigInteger.ONE);
        }
        int T=cin.nextInt();
        for(;T>=1;T--)
        {
            int n;
            BigInteger m;
            n=cin.nextInt();
            m=cin.nextBigInteger();
            for(i=1;i<=1000;i++)
            {
              if(m.compareTo(f[i][0])<1)
              {
                  m=f[i][1].add(m).subtract(BigInteger.ONE);
                  m=m.mod(BigInteger.valueOf(258280327));
                  System.out.println(m);
                  break;
              }
              else
                  m=m.subtract(f[i][0]);
            }
        }
    }
}