有两个正整数,求N!的K进制的位数
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-6;
const double PI = acos(-1.0);
const double e = exp(1.0) ;
int main()
{
long long n,k;
while(cin>>n>>k)
printf("%.0f\n",floor((log(sqrt(2*1.0*PI*k))+k*log(k*1.0/e))/(log(n)))+1);
return 0;
}