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



[align=left]Problem Description[/align]
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]'sLBorder
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|.

[align=left]Input[/align]
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.

[align=left]Output[/align]
The output consists of T
lines. Each has one number, meaning fibn'sLBorderm
modulo 258280327(=2×317+1).

[align=left]Sample Input[/align]

2
4 3
5 5

[align=left]Sample Output[/align]

1
2

[align=left]Author[/align]
SXYZ

[align=left]Source[/align]
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));
}

}
}