#include<stdio.h>

int main(){

int n;

while(1)

{

scanf("%d",&n);

if(n==0){break;}

int a1=0,a2=0,a3=1,b1,b2,b3;

for(int i=2;i<=n;i++){

b1=a1;

b2=a2;

b3=a3;

a1=b3;

a2=b1;

a3=a3+b2;

}

printf("%d\n",a1+a2+a3);

}

return 0;

}

#include<stdio.h> int numb(int n) { int arr[1000] = { 0 }; arr[1] = 1; arr[2] = 2; arr[3] = 3; arr[4] = 4; int i = 4; for (i = 4; i <= n; i++) { arr[i] = arr[i - 1] + arr[i - 3]; } int m = n; if (m <= 4) { return arr[m]; } else { return arr[i-1]; }

}

int main() { int n; while (scanf("%d", &n) &&n!=0) { printf("%d\n", numb(n)); } return 0; }

#include<stdio.h>
int a[1005];
int main()
{
	while(1){
		int t;
		scanf("%d",&t);
		if(!t){
			break;
		}
		a[1]=1,a[2]=2,a[3]=3,a[4]=4;//提前将前四个乌龟数写出
		for(int i=5;i<=t;i++){//从第5个开始当下每个乌龟数均为前一年的加上三年前的
			a[i]=a[i-1]+a[i-3];//因为三年前的所有乌龟在今年都能够生出小乌龟
		}
		printf("%d\n",a[t]);
	}
	return 0;
}
//ps：a[1]为第一年的初始乌龟