数据结构7：递归累加及汉诺塔问题

#include 

//累加函数 
int addTo(int n)
{
	if(n <= 0)
	{
		return 0;
	}
	else
	{
		return addTo(n - 1) + n;
	}
}

//测试
void addToText()
{
	int n;
	
	n = 6;
	printf("0加到%d等于%dn",n,addTo(n));
	
	n = 1;
	printf("0加到%d等于%dn",n,addTo(n));
	
	n = -1;
	printf("0加到%d等于%dn",n,addTo(n));
 } 

int main()
{
	addToText();
 } 

0加到6等于21
0加到1等于1
0加到-1等于0

#include 

//汉诺塔 
void hanoi(int n,char source,char destination,char transit)
{
	if(n <= 0)
		return;
	else
	{
		hanoi(n - 1,source,transit,destination);
		printf("%c -> %c  ",source,destination);
		hanoi(n - 1,transit,destination,source);
	}
}

int hanoiSum(int n)
{
	if(n == 1)
		return 1;
	else
	{
		return 2*hanoiSum(n - 1) + 1;
	}
}

//测试
void hanoiText()
{
	printf("3 plates:rn");
	hanoi(3,'A','B','C');
	printf("n");
	printf("移动次数:%dn",hanoiSum(3));
	
	printf("6 plates:rn");
	hanoi(6,'A','B','C');
	printf("n");
	printf("移动次数:%dn",hanoiSum(6));
	
	printf("1 plates:rn");
	hanoi(1,'A','B','C');
	printf("n");
	printf("移动次数:%dn",hanoiSum(1));
 } 
 
int main()
{
	hanoiText();
}

3 plates:
A -> B  A -> C  B -> C  A -> B  C -> A  C -> B  A -> B
移动次数:7
6 plates:
A -> C  A -> B  C -> B  A -> C  B -> A  B -> C  A -> C  A -> B  C -> B  C -> A  B -> A  C -> B  A -> C  A -> B  C -> B  A -> C  B -> A  B -> C  A -> C  B -> A  C -> B  C -> A  B -> A  B -> C  A -> C  A -> B  C -> B  A -> C  B -> A  B -> C  A -> C  A -> B  C -> B  C -> A  B -> A  C -> B  A -> C  A -> B  C -> B  C -> A  B -> A  B -> C  A -> C  B -> A  C -> B  C -> A  B -> A  C -> B  A -> C  A -> B  C -> B  A -> C  B -> A  B -> C  A -> C  A -> B  C -> B  C -> A  B -> A  C -> B  A -> C  A -> B  C -> B
移动次数:63
1 plates:
A -> B
移动次数:1