HDOJ 5351 MZL's Border 找规律

时间：2015-08-05 18:36:32 阅读：184 评论：0 收藏：0 [点我收藏+]

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打出前i个串的kmp的fail指针:

p: ab
0 0 0 
p: aba
0 0 0 1 
p: abaab
0 0 0 1 1 2 
p: abaababa
0 0 0 1 1 2 3 2 3 
p: abaababaabaab
0 0 0 1 1 2 3 2 3 4 5 6 4 5 
p: abaababaabaababaababa
0 0 0 1 1 2 3 2 3 4 5 6 4 5 6 7 8 9 10 11 7 8 
p: abaababaabaababaababaabaababaabaab
0 0 0 1 1 2 3 2 3 4 5 6 4 5 6 7 8 9 10 11 7 8 9 10 11 12 13 14 15 16 17 18 19 12 13 
p: abaababaabaababaababaabaababaabaababaababaabaababaababa
0 0 0 1 1 2 3 2 3 4 5 6 4 5 6 7 8 9 10 11 7 8 9 10 11 12 13 14 15 16 17 18 19 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 20 21 
p: abaababaabaababaababaabaababaabaababaababaabaababaababaabaababaabaababaababaabaababaabaab
0 0 0 1 1 2 3 2 3 4 5 6 4 5 6 7 8 9 10 11 7 8 9 10 11 12 13 14 15 16 17 18 19 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 33 34

然后就发现规律了.....

MZL‘s Border

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 853 Accepted Submission(s): 275

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

Source

2015 Multi-University Training Contest 5

import java.util.*;
import java.math.*;

public class Main {

	final BigInteger mod = BigInteger.valueOf(258280327);
	
	BigInteger fib[] = new BigInteger[1100];
	BigInteger sum[] = new BigInteger[1100];
	BigInteger all[] = new BigInteger[1100];
	
	void init() {
		
		fib[1]=BigInteger.ONE; fib[2]=BigInteger.ONE;
		sum[1]=BigInteger.ZERO; sum[2]=BigInteger.ONE;
		for(int i=3;i<=1010;i++) {
			fib[i]=fib[i-1].add(fib[i-2]);
			sum[i]=sum[i-1].add(fib[i]);
		}
		all[1]=BigInteger.valueOf(2);
		for(int i=2;i<=1010;i++) {
			all[i]=all[i-1].add(fib[i].multiply(BigInteger.valueOf(2)));
		}
	}
	
	void solve(BigInteger x) {
		if(x.compareTo(BigInteger.valueOf(2))<1) {
			System.out.println(0);
			return ;
		}
		
		for(int i=1;i<=1000;i++) {
			int k=x.compareTo(all[i]);
			if(k==1) continue;
			else if(k==0) {
				System.out.println(sum[i].mod(mod));
				return ;
			}
			else
			{
				BigInteger y = x.subtract(all[i-1]);
				if(y.compareTo(fib[i])>0) {
					y=y.subtract(fib[i]);
				}
				System.out.println(y.add(sum[i-1]).mod(mod));
				return ;
			}
		}
	}
	
	Main(){
		
		init();
		Scanner in = new Scanner(System.in);
		
		int T_T;
		T_T=in.nextInt();
		while(T_T-->0) {
			int n=in.nextInt();
			BigInteger m = in.nextBigInteger();
			solve(m);
		}
	}
	
	public static void main(String[] args) {
		new Main();
	}
}

HDOJ 5351 MZL's Border 找规律

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原文地址：http://blog.csdn.net/ck_boss/article/details/47298991

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