C语言：递归求解F(n)

#include 
int main()
{
	int F[999] , n;
	F[0] = F[1] = 1;
	F[2] = 0;
	printf("请输入要求的项数n:");
	scanf("%d",&n);
	for(int i = 3 ; i <= n ; i++){
		F[i] = F[i-1] - 3*F[i-2] + 2*F[i-3];
	} 
	int flag = 0;
	for(int i = 0 ; i <= n ; i++){
		printf("F[%d] = %dt",i,F[i]);
		flag++;
		if(flag%5==0){
			printf("n");
		}
	}
	int max,min;
	max = min = F[0];
	int count = 0;
	for(int i = 0 ; i <= n ; i++){
		if(F[i] == 0){
			count++;
		}
		if(max < F[i]){
			max = F[i];	
		}else{
			min = F[i];
		}
	}
	printf("为0的项一共有：%dn",count);
	printf("MAX = %d  MIN = %dn",max,min);
	return 0;
}