HDU 1297:Children’s Queue-优快云博客

本文介绍了如何通过编程解决特定的排队问题，满足头目PigHeader提出的规则，即女孩不能单独站在队伍中，或者与另一女孩相邻。使用Java中的BigInteger类型来处理大量数据，实现高效计算。

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Children’s Queue

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 8 Accepted Submission(s) : 1

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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

Author

SmallBeer (CML)

Source

杭电ACM集训队训练赛（VIII）

你离开了，我的世界里只剩下雨。。。

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

public class Main{
    public static void main(String[] args){
        Scanner cin = new Scanner(System.in);
        BigInteger []num = new BigInteger[1010];
        num[1] = BigInteger.valueOf(1);
        num[2] = BigInteger.valueOf(2);
        num[3] = BigInteger.valueOf(4);
        num[4] = BigInteger.valueOf(7);
        for(int i=5; i<1001; i++)
            num[i] = num[i-1].add(num[i-2].add(num[i-4]));
        int n;
        while(cin.hasNext()){
            n = cin.nextInt();
            System.out.println(num[n]);
        }
    }
}

本来想用C/C++做这道题，结果看到(1<=n<=1000)，想到后面估计数据规模非常大，用“__Int64”估计也不行，所以想到了java里面的BigInteger类型，因此就有了上面的代码咯！！

下面这个C/C++语言AC的代码是转载的……

#include<stdio.h>
int f[1001][101];
int main()
{
    int n;
    f[0][1] = 1;
    f[1][1] = 1;
    f[2][1] = 2;
    f[3][1] = 4;
    for(int i = 4; i < 1001; ++i)
        for(int j = 1; j < 101; ++j)
        {
            f[i][j] += f[i - 1][j] + f[i - 2][j] + f[i - 4][j];
            f[i][j + 1] += f[i][j] / 10000;
            f[i][j] %= 10000;
        }
    while(scanf("%d", &n) != EOF)
    {
        int k = 100;
        while(!f[n][k--]);
        printf("%d", f[n][k + 1]);
        for(; k > 0; --k)
            printf("%04d", f[n][k]);
        printf("\n");
    }
    return 0;
}

转载于:https://www.cnblogs.com/im0qianqian/p/5989610.html