《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：

 // 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?
 #include <cstdio>
 #include <vector>
 using namespace std;
 
 int main()
 {
     int n;
     int i;
     vector<int> v;
 
     v.push_back();
     n = ;
     while (true) {
         i = v[n - ];
         if (n >= ) {
             i += v[n - ];
         }
         if (n >= ) {
             i += v[n - ];
         }
         if (i >= ) {
             break;
         } else {
             v.push_back(i);
         }
         ++n;
     }
     printf("n = %d\n", (int)v.size());
 
     while (scanf("%d", &n) ==  && n >  && n < (int)v.size()) {
         printf("%d\n", v[n]);
     }
     v.clear();
 
     return ;
 }