Problem Description
在一无限大的二维平面中,我们做如下假设:
1、 每次只能移动一格; 2、 不能向后走(假设你的目的地是“向上”,那么你可以向左走,可以向右走,也可以向上走,但是不可以向下走); 3、 走过的格子立即塌陷无法再走第二次; 求走n步不同的方案数(2种走法只要有一步不一样,即被认为是不同的方案)。 |
Input
首先给出一个正整数C,表示有C组测试数据
接下来的C行,每行包含一个整数n (n<=20),表示要走n步。 |
Output
请编程输出走n步的不同方案总数;
每组的输出占一行。 |
Sample Input
2 1 2 |
Sample Output
3 7 |
Code
View Code
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Key points
for this question, print out the picture step by step, you will find the regular.
for f(1), you have zero point that you can go two roads, and have one point that you can go three roads.
for f(2), you have two points that you can go two roads, and have one point that you can go three roads.
.....
f(n)=a*2+b*3
a=a+2*b
b=a+b
and firstly a=0, b=1
Also, you can use other formula: f[i]=2*f[i-1]+f[i-2].
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