Question

有这样的数列,1,1,2,3,5,8,13,21……java编程计算第40个数字为多少

Answer 1

前2数相加得出后一个数字，利用递归应可实现

Answer 2

斐波那契数列
代码：
public static void calculate(int first, int second) {
int result = first + second;
System.out.println("1: " + first);
System.out.println("2: " + second);
System.out.println("3: " + result);
//second = result;
for(int i=0; i<37; i++) {
first = second;
second = result;
result = first + second;
System.out.println(i+4 + ": " + result);
}
}

public static void main(String args[]) {
calculate(1,1);
}

结果：
1: 1
2: 1
3: 2
4: 3
5: 5
6: 8
7: 13
8: 21
9: 34
10: 55
11: 89
12: 144
13: 233
14: 377
15: 610
16: 987
17: 1597
18: 2584
19: 4181
20: 6765
21: 10946
22: 17711
23: 28657
24: 46368
25: 75025
26: 121393
27: 196418
28: 317811
29: 514229
30: 832040
31: 1346269
32: 2178309
33: 3524578
34: 5702887
35: 9227465
36: 14930352
37: 24157817
38: 39088169
39: 63245986
40: 102334155

Answer 3

核心代码：
num[0] = 1;
num[1] = 1;
for(int i=2;i<num.length;i++){
num[i]=num[i-1]+num[i-2];
}

结果：

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
最终答案：
102334155

Answer 4

就是简单的递归啊,定义一个参数作为前两项的和,一次循环下去,就可以求出第四十个数了