Children’s Queue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8419    Accepted Submission(s): 2665

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
  
  
  
  

   
    
  
  
  
  

   
   
#include<iostream>
using namespace std;
int f[1005][1201];
int main()
{
	
	int i,j;
	memset(f,sizeof(f));
	f[1][1200]=1;
	f[2][1200]=2;
	f[3][1200]=4;
	f[4][1200]=7;
	//printf("111\n");
	for(i=5;i<=1000;i++)
	{
		for(j=1200;j>=0;j--)
		{
			f[i][j]=f[i-1][j]+f[i-2][j]+f[i-4][j];
		}
		for(j=1200;j>=0;j--)
		{
			if(f[i][j]/10>0)
			{
			    f[i][j-1]+=f[i][j]/10;
				f[i][j]%=10;
			}
		}
	}
	int n;
	while(scanf("%d",&n)>0)
	{
		for(j=0;j<=1200;j++)
			if(f[n][j]!=0)
				break;
			for(;j<=1200;j++)
		printf("%d",f[n][j]);
			printf("\n");
	}
	return 0;
}