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 }