Project Euler 63

题目

Powerful digit counts

The \(5\)-digit number, \(16807=7^5\), is also a fifth power. Similarly, the \(9\)-digit number, \(134217728=8^9\), is a ninth power.

How many \(n\)-digit positive integers exist which are also an \(n^\text{th}\) power?

解决方案

可以发现，\(10^n\)是一个\(n+1\)位数。因此，如果一个数\(a^n\)为\(n\)位数，那么\(a\leq 9\)。

当\(9^n\)的位数小于\(n\)位时，统计结束。（因为\(n\)就算增加\(1\)，\(9^n\)再乘一个\(9\)，也没办法使积的位数增加多于\(1\)位，变成\(n+1\)位）。

代码

from itertools import count

ans = 0
for n in count(1, 1):
    if len(str(9 ** n)) < n:
        break
    for a in range(1, 10):
        if len(str(a ** n)) == n:
            ans += 1
print(ans)