HDUOJ Children’s Queue

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8835    Accepted Submission(s): 2813

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

大数。。 公式 f（x）=f(x-1)+f(x-2)-F（x-4）;

代码：

 1 #include<iostream>

 2 #include<cstdio>

 3 #define maxn 250

 4 #define len 1000

 5 using namespace std;

 6 int a[len+1][maxn+1]={{1},{2},{4},{7}};

 7 int main()

 8 {

 9     int i,j,n,s,c=0;

10     for(i=4;i<=len;i++)

11     {

12         for(c=j=0;j<=maxn;j++)

13         {

14           s=a[i-1][j]+a[i-2][j]+a[i-4][j]+c;

15           a[i][j]=s%10;

16           c=(s-a[i][j])/10;

17         }

18     }

19     while(cin>>n)

20     {

21      for(i=maxn;a[n-1][i]==0;i--);

22      for(j=i;j>=0;j--)

23      {

24          printf("%d",a[n-1][j]);

25      }

26      printf("\n");

27     }

28     return 0;

29 }

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