练习cf1485A. Add and Divide

题目如下
A. Add and Divide
time limit per test1 second
memory limit per test256 megabytes
You have two positive integers 𝑎 and 𝑏.

You can perform two kinds of operations:

𝑎=⌊𝑎/𝑏⌋ (replace 𝑎 with the integer part of the division between 𝑎 and 𝑏)
𝑏=𝑏+1 (increase 𝑏 by 1)
Find the minimum number of operations required to make 𝑎=0.

Input
The first line contains a single integer 𝑡 (1≤𝑡≤100) — the number of test cases.

The only line of the description of each test case contains two integers 𝑎, 𝑏 (1≤𝑎,𝑏≤109).

Output
For each test case, print a single integer: the minimum number of operations required to make 𝑎=0.

题目大意
通过一下两种操作使得a=0，最少需要几次操作。
操作如下1，a = a / b；2. b = b + 1；

题目分析
要使得a = 0，那么最后一步一定是a = a / b，要满足a = 0，那么只要让最后一步的a < b即可。
若初始b=1时必须先对b进行加一的操作。
b越大时a除的越快，可以通过枚举得到效率最高的除数。

点击查看代码

#include <iostream>
using namespace std;

int main(){
    int t; cin >> t;
    while(t--){
        long long a,b;
        cin >> a >> b;
        int ans = 1e9;
        for(int i = 0; i <= 40; i++){
            long long aa = a;
            long long bb = b + i;
            if(bb == 1) {
                continue;
            }
            int cnt = i;
            while(aa >= bb){
                aa /= bb;
                cnt++;
            }
            ans = min(ans,cnt);
        }
        ans++;
        cout << ans << "\n";
    }
    return 0;
}