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：

 #include<iostream>
 #include<cstdio>
 #define maxn 250
 #define len 1000
 using namespace std;
 int a[len+][maxn+]={{},{},{},{}};
 int main()
 {
     int i,j,n,s,c=;
     for(i=;i<=len;i++)
     {
         for(c=j=;j<=maxn;j++)
         {
           s=a[i-][j]+a[i-][j]+a[i-][j]+c;
           a[i][j]=s%;
           c=(s-a[i][j])/;
         }
     }
     while(cin>>n)
     {
      for(i=maxn;a[n-][i]==;i--);
      for(j=i;j>=;j--)
      {
          printf("%d",a[n-][j]);
      }
      printf("\n");
     }
     return ;
 }