hdu1297 - Children’s Queue (递推求解 + 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

杭电ACM集训队训练赛（VIII）

 

Recommend

lcy

 

 

 

                    题意：男女站一排，不能有单个女生自己站的情况，

                            例如：n = 4  则站排情况如下：（F表示女，M表示男）FFFF, FFFM, MFFF, FFMM, MFFM, MMFF, MMMM               结果为7

                                       n = 3  .......................FFM,   FFF,    MMM,    MFF              结果为4

                                       n = 2.........................MM   ,FF                                             结果为2

                                       n = 1.........................M                                                        结果为1

 

                  题解：

                            F(n)表示n个人的合法队列， 根据最后一个人是男是女分为两类：

                            1）最后一人是男人，则对前面n-1个人的队列武任何限制，他站在最后，这种情况共有F(n -1)种

                            2）最后一人是女人，则倒数第二人必须是女生，男生情况则不合法，这种情况又分两种情况：

                                    2.1）如果前F(n - 2)人合法，则加两个女生也合法，所以这种情况共有F(n - 2)种；

                                    2.2)   如果前F(n - 2)人不合法，加上2个女生也可能成为合法队列，这种情况只可能为

                                             F(n - 4) + 男 + 女 这种情况 ，所以这种情况共有F(n - 4)种

                          综上所述，F(n) = F(n - 1) + F(n - 2) + F(n - 4) ，随后就是n<=3情况的特殊除了

                          需要说明的是F(0) = 1根据题意也是合法的，特殊情况

                           

                            

 

 超时的递归代码：现在做这种题如果不打表或是找规律，基本就是超时

 

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

//考察递推求解

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

int fun(int n)
{
	int res;

	if (n == 0)  //特殊处理
	{
		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，但是还是错误，原来是高精度问题，当值为1000时，值大到 __int64 也会溢出

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

//考察递推求解
//本题打表求解


#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代码 ，递推+ 高精度