2015 Multi-University Training Contest 5 1009 MZL's Border

Problem's Link: http://acm.hdu.edu.cn/showproblem.php?pid=5351

[b]Mean:[/b]
给出一个类似斐波那契数列的字符串序列,要你求给出的f
字符串中截取前m位的字符串s中s[1...i] = s[s.size()-i+1....s.size()]的最大长度。

[b]analyse:[/b]

过计算可以发现，b字符串的前缀都是相同的，所以只要求出m的LBorder就行，和n是无关的，打表找出规律，找出对应不同的m，LBorder的值.

[b]Time complexity: O(N)[/b]

[b]Source code: [/b]

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

public class Main {

static BigInteger[] fib1 = new BigInteger[1010];
static BigInteger[] fib2 = new BigInteger[1010];
static BigInteger[] fib3 = new BigInteger[1010];

public static void main(String[] args) {
fib1[1] = new BigInteger("2");
fib1[2] = new BigInteger("2");
fib2[1] = BigInteger.ONE;
fib2[2] = BigInteger.ONE;
fib3[1] = BigInteger.ZERO;
fib3[2] = BigInteger.ONE;

for (int i = 3; i <= 1005; i++) {
fib1[i] = fib1[i - 1].add(fib1[i - 2]);
fib2[i] = fib2[i - 1].add(fib2[i - 2]);
}
for (int i = 3; i <= 1005; i++) {
fib3[i] = fib3[i - 1].add(fib2[i - 1]);
}

Scanner in = new Scanner(System.in);
int t = in.nextInt();
for (int cas = 0; cas < t; ++cas) {
BigInteger m;
int n = in.nextInt(), i;
m = in.nextBigInteger();
for (i = 1; i <= 1005; i++) {
if (m.compareTo(fib1[i]) == 1) {
m = m.subtract(fib1[i]);
} else break;
}
if (m.compareTo(fib1[i].divide(new BigInteger("2"))) == 1) {
m = m.subtract(fib1[i].divide(new BigInteger("2")));
BigInteger ans = m.add(fib3[i]);
ans = ans.subtract(BigInteger.ONE);
ans = ans.remainder(new BigInteger("258280327"));
System.out.println(ans.toString());
} else {
BigInteger ans = m.add(fib3[i]);
ans = ans.subtract(BigInteger.ONE);
ans = ans.remainder(new BigInteger("258280327"));
System.out.println(ans.toString());
}
}
}
}