1 #include "stdafx.h"
 2 #include <string>
 3 using namespace std;
 4 int f(int n)
 5 {
 6     if (n == 0)
 7     {
 8         return 0;
 9     }
10     if (n == 1)
11     {
12         return 1;
13     }
14     return f(n - 1) + f(n - 2);
15 }
16 int _tmain(int argc, _TCHAR* argv[])
17 {
18     printf("%d", f(10));
19     getchar();
20     return 0;
21 }

 1 #include "stdafx.h"
 2 #include <string>
 3 using namespace std;
 4 int f(int a[],int n)
 5 {
 6     if (n == 0)
 7     {
 8         a[0] = 0;
 9         return 0;
10     }
11     if (n == 1)
12     {
13         a[1] = 1;
14         return 1;
15     }
16     if (a[n] >= 0)
17     {
18         return a[n];
19     }
20     a[n] = f(a, n - 1) + f(a, n - 2);
21     return f(a, n - 1) + f(a, n - 2);
22 }
23 int _tmain(int argc, _TCHAR* argv[])
24 {
25     int n = 10;//需要求解的数
26     int* a = (int*)malloc((n + 1)*sizeof(int));
27     for (int i = 0; i < n + 1; i++)
28     {
29         a[i] = -1;
30     }
31     printf("%d\n子问题的解", f(a, n));
32     for (int i = 0; i < n + 1; i++)
33     {
34         printf("%d ", a[i]);
35     }
36     getchar();
37     return 0;
38 }

 1 #include "stdafx.h"
 2 #include <string>
 3 using namespace std;
 4 int f(int a[],int n)
 5 {
 6     a[0] = 0;
 7     a[1] = 1;
 8     for (int i = 2; i <= n; i++)
 9     {
10         a[i] = a[i - 1] + a[i - 2];
11     }
12     return a[n];
13 }
14 int _tmain(int argc, _TCHAR* argv[])
15 {
16     int n = 10;//需要求解的数
17     int* a = (int*)malloc((n + 1)*sizeof(int));
18     for (int i = 0; i < n + 1; i++)
19     {
20         a[i] = -1;
21     }
22     printf("%d\n子问题的解", f(a, n));
23     for (int i = 0; i < n + 1; i++)
24     {
25         printf("%d ", a[i]);
26     }
27     getchar();
28     return 0;
29 }