【HDU 1297】Children’s Queue

 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

【分析】

公式的推导：

1)m

2)mff

3)mfff

a:安全序列后加ff或者m，结果仍然安全。

b:不安全序列后加ff可使其安全，虽然mf加f也能得到安全序列，但与a情况重复。

故：公式a
=a[n-1]+a[n-2]+a[n-4];

另外：还要考虑大数相加的问题：因为：n=1000时输出结果：

12748494904808148294446671041721884239818005733501580815621713101333980596197474

74433619974245291299822523591089179822154130383839594330018972951428262366519975

47955743099808702532134666561848656816661065088789701201682837073071502397487823

19037

【总结】

递推求解分类要全面。

一般假设前n-1个合法，然后找规律。

【参考代码】

/*

ID:smqqlzy

LANG:C

PROB:【HDU 1297】Children’s Queue
*/

#include<stdio.h>

int f[1001][1000];

void add(int sub)

{

    int i;

    for(i=0;i<100;i++)

    {

        f[sub][i]=0;

    }

    for(i=1;i<=f[sub-1][0];i++)

    {

        f[sub][i]+=f[sub-1][i]+f[sub-2][i]+f[sub-4][i];

        f[sub][i+1]=f[sub][i]/10;

        f[sub][i]%=10;

    }

    if(f[sub][f[sub-1][0]+1])

    {

        f[sub][0]=f[sub-1][0]+1;

    }

    else

    {

        f[sub][0]=f[sub-1][0];

    }

}

int main()

{

    int n;

    while(scanf("%d",&n)!=EOF)

    {

        int i;

        for(i=0;i<4;i++)

        {

            f[i][0]=1;

        }

        f[0][1]=1;

        f[1][1]=1;

        f[2][1]=2;

        f[3][1]=4;

        for(i=4;i<=n;i++)

        {

            add(i);

        }

        for(i=f
[0];i>0;i--)

        {

            printf("%d",f
[i]);

        }

        printf("\n");

    }

    return 0;

}