目录
递归和非递归分别实现求第n个斐波那契数
例如:
输入:5 输出:5
输入:10, 输出:55
输入:2, 输出:1
递归实现
#include<stdio.h>
int fib(int n)
{
if (n <= 2)
return 1;
else
return fib(n - 1) + fib(n - 2);
}
int main()
{
int n = 0;
printf("请输入一个数:");
scanf("%d", &n);
int ret = fib(n);
printf("ret=%d\n", ret);
return 0;
}
非递归实现
#include<stdio.h>
int fib(int n)
{
int a = 1;
int b = 1;
int c = 1;//当n=1或2时,返回c=1
while (n >= 3)
{
c = a + b;
a = b;
b = c;
n--;
}
return c;
}
int main()
{
int n = 0;
printf("请输入一个数:");
scanf("%d", &n);
int ret = fib(n);
printf("ret=%d\n", ret);
return 0;
}
运行结果是: