《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(1);
    n = 1;
    while (true) {
        i = v[n - 1];
        if (n >= 2) {
            i += v[n - 2];
        }
        if (n >= 3) {
            i += v[n - 3];
        }
        if (i >= 1000000000) {
            break;
        } else {
            v.push_back(i);
        }
        ++n;
    }
    printf("n = %d
", (int)v.size());
    
    while (scanf("%d", &n) == 1 && n > 0 && n < (int)v.size()) {
        printf("%d
", v[n]);
    }
    v.clear();
    
    return 0;
}