求1-n中x（0-9）的个数

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;
ll count(ll n,ll x){//找1-n中x的个数，时间复杂度log(10,n)
	ll cnt=0,k;
	for (ll i = 1; k = n/i;i*=10){
		ll high=k/10;
		if(x==0){
			if(high) high--;
			else break;
		}
		cnt+=high*i;
		ll cur=k%10;
		if(cur>x) cnt+=i;
		else if(cur==x) cnt+=n-k*i+1;
	}
	return cnt;
}
int main(){
	return 0;
}