Children’s Queue（递推 + JAVA大数）

最新推荐文章于 2022-09-11 12:04:45 发布

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最新推荐文章于 2022-09-11 12:04:45 发布

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分类专栏：大数问题递推文章标签：递推大数问题

本文链接：https://blog.youkuaiyun.com/qq_35104140/article/details/78064623

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本文探讨了一个有趣的儿童队列排列问题，通过分析合法队列的条件，即女生要么不存在要么至少两个相邻，提出了一种递推算法来计算满足条件的不同队列数量。该算法使用了动态规划的思想，并通过预计算的方式提高了效率。

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Children’s Queue
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15331 Accepted Submission(s): 5129

Problem Description
There are many students in PHT School. One day, the headmaster whose name is PigHeader wanted all students stand in a line. He prescribed that girl can not be in single. In other words, either no girl in the queue or more than one girl stands side by side. The case n=4 (n is the number of children) is like
FFFF, FFFM, MFFF, FFMM, MFFM, MMFF, MMMM
Here F stands for a girl and M stands for a boy. The total number of queue satisfied the headmaster’s needs is 7. Can you make a program to find the total number of queue with n children?

Input
There are multiple cases in this problem and ended by the EOF. In each case, there is only one integer n means the number of children (1<=n<=1000)

Output
For each test case, there is only one integer means the number of queue satisfied the headmaster’s needs.

Sample Input

1
2
3

Sample Output

1
2
4

//题目中的意思合法的序列必须为一个女生都没有或者至少有两个女相邻
我们可以假设女生：0 男生：1
假设前n-1位都是一个合法的序列所以当插入第n位时可以分为两种情况
1：当最后一位为男生时此时的种为f[n-1] *1 = f[n-1]
2：当最后一名为女生时那么根据题意第n-1位一定也是一个女生此时又可以分为两种情况
当前n-2为一个合法的序列那么此时的种数为 f[n-2] * 1
当前n-1为一个不和法的序列那么此时序列一定是f[n-4] + 男 + 女所以此时种数为f[n-4]\
所以总的种数为
f[n] = f[n-1] + f[n-4] + f[n-2];

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

public class Main {
    public static void main(String[] args) {

        int n;
        Scanner cin = new Scanner(System.in);
        BigInteger[] qq = new BigInteger[1001];
        qq[1] = BigInteger.ONE;
        qq[2] = qq[1].add(BigInteger.ONE);
        qq[3] = BigInteger.valueOf(4);
        qq[4] = BigInteger.valueOf(7);
        for (int i = 5; i <= 1000; i++)
        {
            qq[i] = qq[i-1].add(qq[i-4]).add(qq[i-2]);
        }
        while (cin.hasNext()) {
            n = cin.nextInt();
            System.out.println(qq[n]);
        }
    }

}