输出斐波那契数列的前n项。斐波那契数列:1 1 2 3 5 8 13 21 .....
时间: 2023-11-18 18:09:54 浏览: 46
以下是Python代码实现:
```python
def fibonacci(n):
a, b = 1, 1
for i in range(n):
print(a, end=' ')
a, b = b, a + b
n = int(input("请输入要输出的斐波那契数列的项数:"))
fibonacci(n)
```
输出:
```
请输入要输出的斐波那契数列的项数:10
1 1 2 3 5 8 13 21 34 55
```
相关问题
1.输出Fibonacci数列:1、1、2、3、5、8、13、21.…的前40项。
Sure, here are the first 40 terms of the Fibonacci sequence:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296.
java输出斐波那契数列的前 20 项:1,1,2,3,5,8,13,21…
可以使用递归或循环来输出斐波那契数列的前20项,以下是使用循环的示例代码:
```java
public class Fibonacci {
public static void main(String[] args) {
int n = 20, t1 = 0, t2 = 1;
System.out.print("斐波那契数列的前 " + n + " 项:");
for (int i = 1; i <= n; ++i) {
System.out.print(t1 + " + ");
int sum = t1 + t2;
t1 = t2;
t2 = sum;
}
}
}
```
输出结果为:斐波那契数列的前 20 项:0 + 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 + 89 + 144 + 233 + 377 + 610 + 987 + 1597 + 2584 + 4181 +