B - Digits /
AtCoder - 5308

Problem Statement
Given is an integer N. Find the number of digits that N has in base K.

Notes
For information on base-K representation, see Positional notation - Wikipedia.

Constraints
All values in input are integers.
1≤N≤109
2≤K≤10
Input
Input is given from Standard Input in the following format:

N K
Output
Print the number of digits that N has in base K.

Sample Input 1
Copy
11 2
Sample Output 1
Copy
4
In binary, 11 is represented as 1011.

Sample Input 2
Copy
1010101 10
Sample Output 2
Copy
7
Sample Input 3
Copy
314159265 3
Sample Output 3
Copy
18

这道题的就是求n的k进制
其实和十进制转二进制差不多
即“除K取余，逆序排列”

附代码：

// 转进制函数
#include<bits/stdc++.h>
using namespace std;
char a[16]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
void t(int n,int k);
int main()
{
    int n,k;
    cin>>n>>k;
    t(n,k);
}
void t(int n,int k){
    int r=n%k;
    n/=k;
    if(n!=0) t(n,k);
    cout<<a[r];
}
\\AC代码：
#include<bits/stdc++.h>
using namespace std;
char a[16]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
int cnt =0;

void t(int n,int k);

int main()
{
    int n,k;
    cin>>n>>k;
    t(n,k);
    cout<<cnt<<endl;
}

void t(int n,int k){
    int r = n%k;
    n /= k;
    if(n!=0) t(n,k);
    cnt++;
}