hdu 1297

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8780    Accepted Submission(s): 2789

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

大数问题：

规律 f(n)=f(n-1)+f(n-2)+f(n-4)

 

#include<stdio.h>

int a[1001][300];

int main()

{

 int i,j,n,t;

 a[1][0]=1;

 a[2][0]=2;

 a[3][0]=4;

 a[4][0]=7;

 for(i=5;i<=1001;i++)//打表

 {  

 t=0;

  for(j=0;j<300;j++)//表示数的位数  

 {   

 a[i][j]=a[i-1][j]+a[i-2][j]+a[i-4][j]+t;  

  t=0;   

 if(a[i][j]>9)   

 {    

 t=a[i][j]/10;  

   a[i][j]%=10;   

 }  

 } 

 }

 while(scanf("%d",&n)!=EOF)

 {   

 for(i=299;i>=0;i--)//忽略前导的0  

  if(a[n][i])   

  break;  

 for(j=i;j>=0;j--)    

 printf("%d",a[n][j]);  

  printf(" ");

 }  

return 0;

}