给定斐波那契数列的项数求对应的数值的普通方法和优化处理

 1 /**
 2  给定斐波那契数列的项数求对应的数值
 3  参考：剑指Offer
 4  */
 5 #include <stdio.h>
 6
 7 int fib(int n);
 8 long long fibonacci(unsigned int n);
 9 int main(int argc, const char * argv[]) {
10
11     int n;
12
13     while (1) {
14
15         printf("请输入你想知道到的斐波那契数列的项数:\t");
16         scanf("%d",&n);
17         int result = fib(n);
18         long long googResult = fibonacci(n);
19         printf("与之对应的斐波那契数列数列的值为:\t%d\t%lld\n\n",result,googResult);
20
21
22     }
23     return 0;
24 }
25
26
27 int fib(int n){
28
29     if(n == 0){
30         return 0;
31     }else if(n == 1){
32         return 1;
33     }else {
34         return fib(n-1) + fib(n-2);
35     }
36
37 }
38
39 //斐波那契数列数列求解优化
40
41 long long fibonacci(unsigned int n){
42
43     int result[] = {0,1};
44     if (n < 2){
45         return result[n];
46     }
47     long long smallest = 0;
48     long long smaller = 1;
49     long long temp = 0;
50
51     for(int i=2; i<=n; i++){
52
53         temp = smaller + smallest;
54
55         smallest = smaller;
56         smaller = temp;
57
58     }
59     return temp;
60
61 }