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  • 483. Smallest Good Base

    For an integer n, we call k>=2 a good base of n, if all digits of n base k are 1.

    Now given a string representing n, you should return the smallest good base of n in string format. 

    Example 1:

    Input: "13"
    Output: "3"
    Explanation: 13 base 3 is 111.
    

    Example 2:

    Input: "4681"
    Output: "8"
    Explanation: 4681 base 8 is 11111.
    

    Example 3:

    Input: "1000000000000000000"
    Output: "999999999999999999"
    Explanation: 1000000000000000000 base 999999999999999999 is 11.
    

    Note:

    1. The range of n is [3, 10^18].
    2. The string representing n is always valid and will not have leading zeros.
     
    Approach #1:
    class Solution {
    public:
        string smallestGoodBase(string n) {
            unsigned long long tn = (unsigned long long)stoll(n);
            unsigned long long x = 1;
            for (int i = 62; i >= 1; --i) {
                if ((x<<i) < tn) {
                    unsigned long long temp = solve(tn, i);
                    if (temp != 0) return to_string(temp);
                }
            }
            return to_string(tn-1);
        }
    private:
        unsigned long long solve(unsigned long long num, int d) {
            double tn = (double) num;
            unsigned long long r = (unsigned long long)(pow(tn, 1.0/d)+1);
            unsigned long long l = 1;
            while (l <= r) {
                unsigned long long sum = 1;
                unsigned long long cur = 1;
                unsigned long long m = l + (r - l) / 2;
                for (int i = 1; i <= d; ++i) {
                    cur *= m;
                    sum += cur;
                }
                if (sum == num) return m;
                if (sum < num) l = m + 1;
                else r = m - 1;
            }
            return 0;
        }
    };
    
    Runtime: 4 ms, faster than 49.59% of C++ online submissions for Smallest Good Base.

    come from: https://leetcode.com/problems/smallest-good-base/discuss/96590/3ms-AC-C%2B%2B-long-long-int-%2B-binary-search

    永远渴望,大智若愚(stay hungry, stay foolish)
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/9925875.html
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