转载请注明出处： http://www.cnblogs.com/fraud/          ——by fraud

MZL‘s Border

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 905    Accepted Submission(s): 295

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

  1 import java.io.OutputStream;
  2 import java.io.IOException;
  3 import java.io.InputStream;
  4 import java.io.PrintWriter;
  5 import java.util.StringTokenizer;
  6 import java.math.BigInteger;
  7 import java.io.IOException;
  8 import java.io.BufferedReader;
  9 import java.io.InputStreamReader;
 10 import java.io.InputStream;
 11
 12 /**
 13  * Built using CHelper plug-in
 14  * Actual solution is at the top
 15  *
 16  * @author [email protected]/fraud
 17  */
 18 public class Main {
 19     public static void main(String[] args) {
 20         InputStream inputStream = System.in;
 21         OutputStream outputStream = System.out;
 22         Scanner in = new Scanner(inputStream);
 23         PrintWriter out = new PrintWriter(outputStream);
 24         Task1009 solver = new Task1009();
 25         solver.solve(1, in, out);
 26         out.close();
 27     }
 28
 29     static class Task1009 {
 30         Scanner in;
 31         PrintWriter out;
 32
 33         public void solve(int testNumber, Scanner in, PrintWriter out) {
 34             this.in = in;
 35             this.out = out;
 36             run();
 37         }
 38
 39         void run() {
 40             BigInteger dp[] = new BigInteger[2010];
 41             BigInteger a[] = new BigInteger[2010];
 42             dp[0] = BigInteger.ONE;
 43             dp[1] = BigInteger.ONE;
 44             dp[2] = BigInteger.ONE;
 45             dp[3] = BigInteger.valueOf(3);
 46             dp[4] = BigInteger.valueOf(5);
 47             a[0] = BigInteger.ZERO;
 48             a[1] = BigInteger.ZERO;
 49             a[2] = BigInteger.ONE;
 50             a[3] = BigInteger.ONE;
 51             a[4] = BigInteger.valueOf(2);
 52             for (int i = 5; i < 2010; i++) {
 53                 dp[i] = dp[i - 1].add(dp[i - 2]);
 54                 a[i] = a[i - 2].add(dp[i - 2]);
 55             }
 56             for (int i = 1; i < 2010; i++) {
 57                 dp[i] = dp[i].add(dp[i - 1]);
 58             }
 59             BigInteger m;
 60             int t, n;
 61             t = in.nextInt();
 62             while (t != 0) {
 63                 t--;
 64                 n = in.nextInt();
 65                 m = in.nextBigInteger();
 66                 if (m.compareTo(BigInteger.ONE) == 0) {
 67                     out.println(1);
 68                     continue;
 69                 }
 70                 int i = 0;
 71                 for (i = 0; i < 2010; i++) {
 72                     if (dp[i].compareTo(m) >= 0) break;
 73                 }
 74                 i--;
 75                 out.println(a[i + 1].add(m.subtract(dp[i].add(BigInteger.ONE))).mod(BigInteger.valueOf(258280327)));
 76             }
 77         }
 78
 79     }
 80
 81     static class Scanner {
 82         BufferedReader br;
 83         StringTokenizer st;
 84
 85         public Scanner(InputStream in) {
 86             br = new BufferedReader(new InputStreamReader(in));
 87             eat("");
 88         }
 89
 90         private void eat(String s) {
 91             st = new StringTokenizer(s);
 92         }
 93
 94         public String nextLine() {
 95             try {
 96                 return br.readLine();
 97             } catch (IOException e) {
 98                 return null;
 99             }
100         }
101
102         public boolean hasNext() {
103             while (!st.hasMoreTokens()) {
104                 String s = nextLine();
105                 if (s == null)
106                     return false;
107                 eat(s);
108             }
109             return true;
110         }
111
112         public String next() {
113             hasNext();
114             return st.nextToken();
115         }
116
117         public int nextInt() {
118             return Integer.parseInt(next());
119         }
120
121         public BigInteger nextBigInteger() {
122             return new BigInteger(next());
123         }
124
125     }
126 }

hdu5351 MZL's Border(规律题，java)