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  • 多校5-MZL's Border 分类: 比赛 2015-08-05 21:28 7人阅读 评论(0) 收藏

    MZL’s Border
    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 909 Accepted Submission(s): 297

    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

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    import java.math.*;
    import java.util.*;
    
    public class Main {
        public static void main(String[] args)
        {
            Scanner in = new Scanner (System.in);
            int i;
            BigInteger[] a = new BigInteger[1015];
            int t=in.nextInt();
            BigInteger ans;
            a[1] = BigInteger.valueOf(1);
            a[2] = BigInteger.valueOf(2);
            for(i=3;i<=1000;i++)
            {
                a[i]=a[i-1].add(a[i-2]);
            }
            while(t>0)
            {
                t--;
                int n=in.nextInt();
                BigInteger m = in.nextBigInteger();
                for(i=1;i<=1000;i++)
                {
                    if(m.compareTo(a[i])<0&&m.compareTo(a[i-1])>=0)
                    {
                        ans = m.subtract(a[i-2]).mod(BigInteger.valueOf(258280327));
                        System.out.println(ans);
                        break;
                    }
    
                }
            }
        }
    }
    

    版权声明:本文为博主原创文章,未经博主允许不得转载。

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  • 原文地址:https://www.cnblogs.com/juechen/p/4721926.html
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