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HDU 5351 MZL's Border(规律)

alonwang 2023-05-04 阅读 29


MZL's Border


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 835    Accepted Submission(s): 268



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


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


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


 




点击打开链接






import java.util.Scanner;
import java.math.BigInteger;

public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        BigInteger[] a=new BigInteger[1055];
        BigInteger[] b=new BigInteger[1055];
        a[1]=BigInteger.valueOf(1);
        a[2]=BigInteger.valueOf(2);
        for(int i=3;i<1050;++i)
        {
            a[i]=a[i-1].add(a[i-2]);
        }
        while(N>0)
        {
            N--;
            int n=sc.nextInt();
            BigInteger m = sc.nextBigInteger();

            for(int i=1;i<1050;++i)
            {
                if( m.compareTo(a[i+1].subtract(BigInteger.valueOf(1)))<0 && m.compareTo(a[i].subtract(BigInteger.valueOf(1)))>=0 )
                {

                     System.out.println((m.subtract(a[i-1])).mod(BigInteger.valueOf(258280327)));
                     break;
                }

            }
        }
    }
}




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