Project Euler 113

题目

Non-bouncy numbers

Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, \(134468\).

Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, \(66420\).

We shall call a positive integer that is neither increasing nor decreasing a "bouncy" number; for example, \(155349\).

As \(n\) increases, the proportion of bouncy numbers below n increases such that there are only \(12951\) numbers below one-million that are not bouncy and only \(277032\) non-bouncy numbers below \(10^{10}\).

How many numbers below a googol (\(10^{100}\)) are not bouncy?

解决方案

所谓的上升数和下降数，下一位的值只能都大于等于上一位或者小于等于上一位。因此可以用动态规划的方法统计\(n\)位数的上升数和下降数。

分别设两个状态\(f(i,d),g(i,d)(1\le i\le n,0\le d\le 9)\)表示\(i\)位数中，个位为\(d\)的上升数/下降数分别有多少个。

注意到上升数的每一位中，下一位的值只能都大于等于上一位，可以列出\(f(i,j)\)的状态转移方程：

\[ f(i,d)= \left \{\begin{aligned} &0 & & \text{if}\quad i=1\land d=0 \\ &1 & & \text{else if}\quad i=1 \\ &\sum_{j=0}^d f(i-1,j) & & \text{else} \end{aligned}\right. \]

类似的，可以列出\(g(i,j)\)的状态转移方程：

\[ g(i,d)= \left \{\begin{aligned} &0 & & \text{if}\quad i=1\land d=0 \\ &1 & & \text{else if}\quad i=1 \\ &\sum_{j=d}^9 g(i-1,j) & & \text{else} \end{aligned}\right. \]

注意到，\(111\dots111,222\dots222\)这种\(n\)位都是一样的数都算进了上升数和下降数中，因此最终答案需要减去这两种数多算的一次。

最终答案为：

\[\sum_{i=1}^n\sum_{d=0}^9(f(i,d)+g(i,d)-1)\]

代码

N = 100
f = [1 for _ in range(10)]
f[0] = 0
g = f.copy()
ans = 9
for i in range(N - 1):
    f = [sum(f[:i + 1]) for i in range(10)]
    g = [sum(g[i:]) for i in range(10)]
    ans += sum(f) + sum(g) - 9
print(ans)