设 G 是一个有向图，其顶点集是从 1 到 100 的数字集合。当满足 j = i + 1 或 j = 3i 时，从顶点 i 到顶点 j 有一条边。 G 中从顶点 1 到顶点 100 的路径中的最小边数是
(一) 4
(乙) 7
(三) 23
(四) 99答案：(乙)
说明：任务是在 G 中从顶点 1 到顶点 100 的路径中找到最少边数，以便我们可以从顶点 i 移动到 i+1 或 3i。
Since the task is to minimize number of edges,
we would prefer to follow 3*i.
Let us follow multiple of 3.
1 => 3 => 9 => 27 => 81, now we can't follow multiple
of 3. So we will have to follow i+1. This solution gives
a long path.
What if we begin from end, and we reduce by 1 if
the value is not multiple of 3, else we divide by 3.
100 => 99 => 33 => 11 => 10 => 9 => 3 => 1
So we need total 7 edges.这个问题的测验