hdu1297 - Children’s Queue (遞推求解 + 高精度)

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10272    Accepted Submission(s): 3289

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

杭電ACM集訓隊訓練賽（VIII）

 

Recommend

lcy

 

 

 

                    題意：男女站一排，不能有單個女生自己站的情況，

                            例如：n = 4  則站排情況如下：（F表示女，M表示男）FFFF, FFFM, MFFF, FFMM, MFFM, MMFF, MMMM               結果爲7

                                       n = 3  .......................FFM,   FFF,    MMM,    MFF              結果爲4

                                       n = 2.........................MM   ,FF                                             結果爲2

                                       n = 1.........................M                                                        結果爲1

 

                  題解：

                            F(n)表示n個人的合法隊列， 根據最後一個人是男是女分爲兩類：

                            1）最後一人是男人，則對前面n-1個人的隊列武任何限制，他站在最後，這種情況共有F(n -1)種

                            2）最後一人是女人，則倒數第二人必須是女生，男生情況則不合法，這種情況又分兩種情況：

                                    2.1）如果前F(n - 2)人合法，則加兩個女生也合法，所以這種情況共有F(n - 2)種；

                                    2.2)   如果前F(n - 2)人不合法，加上2個女生也可能成爲合法隊列，這種情況只可能爲

                                             F(n - 4) + 男 + 女 這種情況 ，所以這種情況共有F(n - 4)種

                          綜上所述，F(n) = F(n - 1) + F(n - 2) + F(n - 4) ，隨後就是n<=3情況的特殊除了

                          需要說明的是F(0) = 1根據題意也是合法的，特殊情況

                           

                            

 

 超時的遞歸代碼：現在做這種題如果不打表或是找規律，基本就是超時

 

/******************************
*
*	acm:   hdu-1297
*
*	title: Children’s Queue
*
*	time:  2014.5.18
*
******************************/

//考察遞推求解

#include <stdio.h>
#include <stdlib.h>

int fun(int n)
{
	int res;

	if (n == 0)  //特殊處理
	{
		res = 1;
	}
	else if (n == 1)
	{
		res = 1;
	}
	else if (n == 2)
	{
		res = 2;
	}
	else if (n == 3)
	{
		res = 4;
	}
	else
	{
		res = fun(n-1) + fun(n-2) + fun(n-4);
	}

	return res;
}

int main()
{
	int n;

    while (~scanf("%d", &n))
	{
		printf("%d\n", fun(n));
	}

    return 0;
}

 

 

//我用打表方法做，但是還是錯誤，原來是高精度問題，當值爲1000時，值大到 __int64 也會溢出

/******************************
*
*	acm:   hdu-1297
*
*	title: Children’s Queue
*
*	time:  2014.5.18
*
******************************/

//考察遞推求解
//本題打表求解


#include <stdio.h>
#include <stdlib.h>

#define ll __int64

ll Array[1000];

void fun(int n)
{
	int i;

	Array[0] = 1;
	Array[1] = 1;
	Array[2] = 2;
	Array[3] = 4;

	for (i = 4; i < n; i++)
	{
		Array[i] = Array[i-1] + Array[i-2] + Array[i-4];
	}
}

int main()
{
	int n;

	fun(1001);

    while (~scanf("%d", &n))
	{
		printf("%I64d\n", Array[n]);
	}

    return 0;
}

 

//AC代碼 ，遞推+ 高精度