《Cracking the Coding Interview》——第9章：递归和动态规划——题目1

2014-03-20 02:55

题目：小朋友跳台阶，每次跳1层或2层，那么跳N层总共有多少种跳法。

解法：斐波那契数列。

代码：

 1 // 9.1 A child can run up the stair with n staircases. Every time he can hop up by 1, 2 or 3 steps. How many possible way to do this are there?

 2 #include <cstdio>

 3 #include <vector>

 4 using namespace std;

 5 

 6 int main()

 7 {

 8     int n;

 9     int i;

10     vector<int> v;

11     

12     v.push_back(1);

13     n = 1;

14     while (true) {

15         i = v[n - 1];

16         if (n >= 2) {

17             i += v[n - 2];

18         }

19         if (n >= 3) {

20             i += v[n - 3];

21         }

22         if (i >= 1000000000) {

23             break;

24         } else {

25             v.push_back(i);

26         }

27         ++n;

28     }

29     printf("n = %d\n", (int)v.size());

30     

31     while (scanf("%d", &n) == 1 && n > 0 && n < (int)v.size()) {

32         printf("%d\n", v[n]);

33     }

34     v.clear();

35     

36     return 0;

37 }