斐波那契数列raptor
时间: 2023-12-27 08:25:01 浏览: 211
以下是使用Raptor进行斐波那契数列的演示:
```python
# Raptor代码
N = input("请输入所需前几位数:")
N = int(N)
a = 1
b = 1
print("将显示斐波那契数列前{}位".format(N))
print("i=1")
print(a)
print("i=2")
print(b)
i = 3
while i <= N:
print("i={}".format(i))
c = a + b
print(c)
a = b
b = c
i += 1
```
相关问题
raptor斐波那契数列
斐波那契数列是指从第三项开始,每一项都等于前两项之和。在代码中,我们可以使用循环语句来实现斐波那契数列的计算。以下是一个使用while循环的C语言代码示例:
#include <stdio.h>
int main() {
int a = 1, b = 1, c, i, N;
printf("请输入所需前几位数\n");
scanf("%d", &N);
printf("将显示斐波那契数列前%d位\n", N);
printf("i=1\n");
printf("%d\n", a);
printf("i=2\n");
printf("%d\n", b);
i = 3;
while (i <= N) {
printf("i=%d\n", i);
c = a + b;
printf("%d\n", c);
a = b;
b = c;
++i;
}
return 0;
}
在这个代码中,我们首先定义了三个变量a、b和c,分别表示斐波那契数列中的前两项和当前项。然后,我们通过scanf函数从用户输入中获取需要计算的斐波那契数列的项数N。接下来,我们使用while循环来计算斐波那契数列,并使用printf函数输出每一项的值。
raptor求斐波那契数列
斐波那契数列是一个数列,其中每个数字是前两个数字之和。根据给定的代码,可以使用集合、和递归三种方法来求解斐波那契数列。
使用集合方法:
```java
import java.util.ArrayList;
public class Main {
public static void main(String[] args) {
ArrayList<Integer> list = new ArrayList<>();
int count = 0;
list.add(1);
list.add(1);
for (int i = 0; i < 18; i++) {
list.add(list.get(i) + list.get(i + 1));
}
for (int a : list) {
count += a;
}
System.out.println("前20项的和为:" + count);
System.out.println("第20项的值为:" + list.get(19));
}
}
```
使用数组方法:
```java
public class Main {
public static void main(String[] args) {
int[] array = new int[20];
int count = 0;
array[0] = 1;
array[1] = 1;
for (int i = 2; i < 20; i++) {
array[i] = array[i - 1] + array[i - 2];
}
for (int a : array) {
count += a;
}
System.out.println("前20项的和为:" + count);
System.out.println("第20项的值为:" + array[19]);
}
}
```
使用递归方法:
```java
public class Main {
public static void main(String[] args) {
int count = 0;
for (int i = 1; i <= 20; i++) {
count += number(i);
}
System.out.println("前20项的和为:" + count);
System.out.println("第20项的值为:" + number(20));
}
public static int number(int num) {
if (num == 1 || num == 2) {
return 1;
} else {
return number(num - 2) + number(num - 1);
}
}
}
```
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