2
4
5
0 

Sample Output

2
4
6 

#include<stdio.h>
int main()
{
	int n, i, count, f[100];
	while(scanf("%d",&n)==1)
	{
		count=0;
		if(n==0)
		{
			break;
		}
		if(n<=3)
		{
			count=n;
		}
		else
		{
			for(i=4;i<=n;i++)
			{
				f[1]=1;
				f[2]=2;
				f[3]=3;
				f[i]=f[i-1]+f[i-3];//<span style="font-family:Verdana;">斐波那契数列的通项公式的变形</span>
			}
			count=f[n];
		}
		printf("%d\n",count);
	}
	return 0;
}