Recursive Algorithm 8-The Step Problem of Complex Recursion

If someone goes up the stairs, one step can buy one step, two steps or three steps, a total of n steps. It is programmed to output all possible methods. For example, n=4

1    1    1    1
1    1    2
1    2    1
1    3
2    1    1
2    2
3    1

According to the meaning of the question, the question can be divided into three situations: one step at a time, two steps at a time, and three steps at a time. In the recursive function, a parameter n needs to be introduced to indicate how many steps are taken each time. Function prototype:

void step(int n)

Use the array queue to store the number of stages on each stage. If you go up a step, store 1 in the array queue, the code is as follows:

queue[index++]=1;
step(n-1);
--index;

If you go up two steps, store 2 in the array queue, the code is as follows:
if(n>1)
{     queue[index++]=2;     step(n-2);     --index; }

If you go up three steps, store 3 into the array queue, the code:
if(n>2)
{     queue[index++]=3;     step(n-3);     --index; }

When n=0, output the method of each step up.

Code:

#include<stdio.h>
#include <iostream>
void output();
void step(int n);
#define STAIR_NUM 4
int queue[STAIR_NUM];
int total = 0;
int index;
int main()
{
	int step_num=4;

	printf("--------------------------------------\n");
	printf("             上台阶的方法               \n");
	printf("--------------------------------------\n");
	step(step_num);
	printf("\n 共有 %d 种方法 \n", total);
	getchar();
}
void step(int n)
{
	if (n == 0)
	{
		total++;
		printf("---------第%d种方法 ----------\n", total);
		output();
		return;
	}
	queue[index++] = 1;
	step(n - 1);
	--index;
	if (n > 1)
	{
		queue[index++] = 2;
		step(n - 2);
		--index;
	}
	if (n > 2)
	{
		queue[index++] = 3;
		step(n - 3);
		--index;
	}
}
void output()
{
	int i;
	for (i = 0; i < index; i++)
		printf("-%d", queue[i]);
	printf("-\n");
}

result: