HDU 5351(MZL's Border-Java的BigInteger类)

MZL's Border

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1102    Accepted Submission(s): 353

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 i s 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
   
   
   
   

 

Author

SXYZ

 

Source

2015 Multi-University Training Contest 5

 

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import java.io.*; 
import java.util.*; 
import java.math.*;
public class Main{
	static BigInteger f[] = new BigInteger[2000];
	static BigInteger F = new BigInteger("258280327");
    public static void main(String[] arg){
    	Scanner cin = new Scanner(System.in);
    	
    	f[2] = f[1]= new BigInteger("1");
    	for(int i=3;i<=1000;i++) f[i]=f[i-1].add(f[i-2]);
    	int T=cin.nextInt();
    	while ((T--)>0) {
    		int n=cin.nextInt();
    		BigInteger m = cin.nextBigInteger();
    		m=m.add(BigInteger.ONE);
    		int i=1;
    		while (m.compareTo(f[i])>=0) ++i;
    		System.out.println((m.subtract(BigInteger.ONE).subtract(f[i-2])).mod(F));
    	}
    	
    }
}