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feat: add solutions to lc problem: No.0483
No.0483.Smallest Good Base
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‎solution/0400-0499/0483.Smallest Good Base/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：数学**
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假设 $n$ 在 $k$ 进制下的所有位数均为 $1$，且位数为 $m+1$，那么有式子 ①：
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$$
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n=k^0+k^1+k^2+...+k^m
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$$
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当 $m=0$ 时，上式 $n=1$，而题目 $n$ 取值范围为 $[3, 10^{18}]$，因此 $m>0$。
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当 $m=1$ 时，上式 $n=k^0+k^1=1+k$，即 $k=n-1>=2$。
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我们来证明一般情况下的两个结论，以帮助解决本题。
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**结论一：** $m<\log _{k} n$
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注意到式子 ① 是个首项为 $1$，且公比为 $k$ 的等比数列。利用等比数列求和公式，我们可以得出：
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$$
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n=\frac{1-k^{m+1}}{1-k}
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$$
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变形得：
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$$
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k^{m+1}=k \times n-n+1 < k \times n
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$$
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移项得：
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$$
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m<\log _{k} n
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$$
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题目 $n$ 取值范围为 $[3, 10^{18}]$，又因为 $k>=2$，因此 $m<\log _{k} n<\log _{2} 10^{18}<60$。
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**结论二：** $k=\left \lfloor \sqrt[m]{n} \right \rfloor $
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$$
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n=k^0+k^1+k^2+...+k^m>k^m
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$$
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根据二项式定理：
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$$
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(a+b)^{n}=\sum_{k=0}^{n}\left(\begin{array}{l}
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n \\
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k
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\end{array}\right) a^{n-k} b^{k}
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$$
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整合，可得：
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$$
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(k+1)^{m}=\left(\begin{array}{c}
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m \\
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0
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\end{array}\right) k^{0}+\left(\begin{array}{c}
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m \\
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1
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\end{array}\right) k^{1}+\left(\begin{array}{c}
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m \\
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2
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\end{array}\right) k^{2}+\cdots+\left(\begin{array}{c}
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m \\
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m
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\end{array}\right) k^{m}
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$$
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当 $m>1$ 时，满足：
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$$
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\forall i \in[1, m-1],\left(\begin{array}{c}
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m \\
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i
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\end{array}\right)>1
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$$
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所以有：
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$$
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\begin{aligned}
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(k+1)^{m} &=\left(\begin{array}{c}
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m \\
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0
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\end{array}\right) k^{0}+\left(\begin{array}{c}
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m \\
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1
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\end{array}\right) k^{1}+\left(\begin{array}{c}
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m \\
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2
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\end{array}\right) k^{2}+\cdots+\left(\begin{array}{c}
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m \\
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m
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\end{array}\right) k^{m} \\
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&>k^{0}+k^{1}+k^{2}+\cdots+k^{m}=n
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\end{aligned}
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$$
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即：
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$$
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k < \sqrt[m]{n} < k+1
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$$
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由于 $k$ 是整数，因此 $k=\left \lfloor \sqrt[m]{n} \right \rfloor $。
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综上，依据结论一，我们知道 $m$ 的取值范围为 $[1,log_{k}n)$，且 $m=1$ 时必然有解。随着 $m$ 的增大，进制 $k$ 不断减小。所以我们只需要从大到小检查每一个 $m$ 可能的取值，利用结论二快速算出对应的 $k$ 值，然后校验计算出的 $k$ 值是否有效即可。如果 $k$ 值有效，我们即可返回结果。
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时间复杂度 $O(log^{2}n)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def smallestGoodBase(self, n: str) -> str:
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def cal(k, m):
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p = s = 1
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for i in range(m):
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p *= k
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s += p
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return s
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num = int(n)
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for m in range(63, 1, -1):
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l, r = 2, num - 1
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while l < r:
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mid = (l + r) >> 1
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if cal(mid, m) >= num:
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r = mid
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else:
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l = mid + 1
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if cal(l, m) == num:
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return str(l)
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return str(num - 1)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public String smallestGoodBase(String n) {
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long num = Long.parseLong(n);
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for (int len = 63; len >= 2; --len) {
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long radix = getRadix(len, num);
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if (radix != -1) {
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return String.valueOf(radix);
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}
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}
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return String.valueOf(num - 1);
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}
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private long getRadix(int len, long num) {
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long l = 2, r = num - 1;
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while (l < r) {
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long mid = l + r >>> 1;
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if (calc(mid, len) >= num) r = mid;
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else l = mid + 1;
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}
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return calc(r, len) == num ? r : -1;
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}
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private long calc(long radix, int len) {
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long p = 1;
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long sum = 0;
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for (int i = 0; i < len; ++i) {
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if (Long.MAX_VALUE - sum < p) {
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return Long.MAX_VALUE;
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}
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sum += p;
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if (Long.MAX_VALUE / p < radix) {
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p = Long.MAX_VALUE;
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} else {
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p *= radix;
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}
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}
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return sum;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string smallestGoodBase(string n) {
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long v = stol(n);
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int mx = floor(log(v) / log(2));
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for (int m = mx; m > 1; --m) {
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int k = pow(v, 1.0 / m);
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long mul = 1, s = 1;
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for (int i = 0; i < m; ++i) {
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mul *= k;
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s += mul;
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}
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if (s == v) {
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return to_string(k);
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}
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}
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return to_string(v - 1);
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}
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};
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```
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### **...**

‎solution/0400-0499/0483.Smallest Good Base/README_EN.md

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### **Python3**
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```python
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class Solution:
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def smallestGoodBase(self, n: str) -> str:
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def cal(k, m):
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p = s = 1
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for i in range(m):
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p *= k
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s += p
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return s
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num = int(n)
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for m in range(63, 1, -1):
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l, r = 2, num - 1
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while l < r:
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mid = (l + r) >> 1
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if cal(mid, m) >= num:
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r = mid
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else:
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l = mid + 1
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if cal(l, m) == num:
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return str(l)
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return str(num - 1)
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```
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### **Java**
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```java
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class Solution {
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public String smallestGoodBase(String n) {
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long num = Long.parseLong(n);
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for (int len = 63; len >= 2; --len) {
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long radix = getRadix(len, num);
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if (radix != -1) {
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return String.valueOf(radix);
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}
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}
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return String.valueOf(num - 1);
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}
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private long getRadix(int len, long num) {
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long l = 2, r = num - 1;
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while (l < r) {
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long mid = l + r >>> 1;
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if (calc(mid, len) >= num) r = mid;
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else l = mid + 1;
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}
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return calc(r, len) == num ? r : -1;
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}
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private long calc(long radix, int len) {
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long p = 1;
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long sum = 0;
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for (int i = 0; i < len; ++i) {
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if (Long.MAX_VALUE - sum < p) {
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return Long.MAX_VALUE;
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}
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sum += p;
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if (Long.MAX_VALUE / p < radix) {
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p = Long.MAX_VALUE;
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} else {
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p *= radix;
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}
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}
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return sum;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string smallestGoodBase(string n) {
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long v = stol(n);
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int mx = floor(log(v) / log(2));
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for (int m = mx; m > 1; --m) {
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int k = pow(v, 1.0 / m);
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long mul = 1, s = 1;
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for (int i = 0; i < m; ++i) {
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mul *= k;
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s += mul;
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}
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if (s == v) {
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return to_string(k);
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}
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}
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return to_string(v - 1);
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}
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};
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```
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### **...**

‎solution/0400-0499/0483.Smallest Good Base/Solution.cpp

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class Solution {
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public:
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string smallestGoodBase(string n) {
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long v = stol(n);
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int mx = floor(log(v) / log(2));
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for (int m = mx; m > 1; --m) {
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int k = pow(v, 1.0 / m);
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long mul = 1, s = 1;
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for (int i = 0; i < m; ++i) {
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mul *= k;
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s += mul;
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}
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if (s == v) {
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return to_string(k);
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}
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}
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return to_string(v - 1);
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}
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};

‎solution/0400-0499/0483.Smallest Good Base/Solution.py

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class Solution:
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def smallestGoodBase(self, n: str) -> str:
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def cal(k, m):
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p = s = 1
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for i in range(m):
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p *= k
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s += p
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return s
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num = int(n)
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for m in range(63, 1, -1):
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l, r = 2, num - 1
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while l < r:
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mid = (l + r) >> 1
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if cal(mid, m) >= num:
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r = mid
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else:
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l = mid + 1
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if cal(l, m) == num:
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return str(l)
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return str(num - 1)

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