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+1][maxn+1]={{1},{2},{4},{7}};
int main()
{
    int i,j,n,s,c=0;
    for(i=4;i<=len;i++)
    {
        for(c=j=0;j<=maxn;j++)
        {
          s=a[i-1][j]+a[i-2][j]+a[i-4][j]+c;
          a[i][j]=s%10;
          c=(s-a[i][j])/10;
        }
    }
    while(cin>>n)
    {
     for(i=maxn;a[n-1][i]==0;i--);
     for(j=i;j>=0;j--)
     {
         printf("%d",a[n-1][j]);
     }
     printf("
");
    }
    return 0;
}