C语言第11天，函数递归（二）

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>

//求n的阶乘
int Fac1(int n) {
	int i = 0;
	int ret = 1;
	for (i = 1; i <= n; i++) {
		ret *= i;
	}
	return ret;
}
//递归的方式
int Fac2(int n) {
	if (n <= 1)
		return 1;
	else
		return n * Fac2(n - 1);

}

int main() {
	//求n的阶乘
	int n = 0;
	int ret = 0;
	int ret2 = 0;
	scanf("%d", &n);
	ret = Fac1(n);//循环的方式
	ret2 = Fac2(n);
	printf("%d\n", ret);
	printf("--------------------------\n");
	printf("%d\n", ret2);
	
	return 0;
}

5
120
--------------------------
120

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>

//斐波那契数列
//1 1 2 3 5 8 13 21 34 55......

//用递归反而做了大量重复运算
int count = 0;
int Fib(int n) {
	if (n == 3) {//测试第3个斐波那契数的计算次数
		count++;
	}
	if (n <= 2)
		return 1;
	else
		return Fib(n - 1) + Fib(n - 2);
}

int Fib2(int n) {
	int a = 1;
	int b = 1;
	int c = 1;
	while (n>2)
	{
		c = a + b;
		a = b;
		b = c;
		n--;
	}
	return c;
}

int main() {

	int n = 0;
	int ret = 0;
	int ret2 = 0;
	scanf("%d",&n);
	ret = Fib(n);
	printf("ret = %d\n", ret);
	printf("count = %d\n", count);

	printf("-----------------------------\n");

	ret2 = Fib2(n);
	printf("ret2 = %d\n", ret2);
	return 0;
}

10
ret = 55
count = 21
-----------------------------
ret2 = 55