How to generate the nth number that consists only of even digits? - c++

Question

I'm novice to Programming. I can find numbers consisting of even digits but my algorithm complexity is O(n). For large n my algorithm is too slow. So I need a more efficient algorithm. Can anyone help me?
For example, the first numbers with even digits are 0 , 2 , 4 , 6 , 8 , 20 , 22 , 24 , 26 , 28 , 40 etc. 2686 is another example of a number with even digits.
Here is my code: http://ideone.com/nsBzej
#include<bits/stdc++.h>
using namespace std;
long long int a[10],b[20];
long long int powr(int i)
{
long long int ans=5;
for(int j=2;j<=i;j++)
{
ans=ans*5;
}
return ans;
}
int main()
{
//freopen("input.txt","r",stdin);
//freopen("output.txt","w",stdout);
long long int n,s,sum,p;
int t;
cin>>t;
for(int j=1;j<=t;j++)
{
s=20,sum=0;
a[1]=0, a[2]=2, a[3]=4, a[4]=6, a[5]=8;
for(int i=1;i<=17;i++)
{
b[i]=s;
s=s*10;
}
cin>>n;
for(int i=17;i>=1;i--)
{
p=powr(i);
while(p<n)
{
sum=sum+b[i];
n=n-p;
}
}
printf("Case %d: %lld\n",j,sum);
}
}
It is complexity O(n). But I get wrong verdict.

Answer 1

#include <stdio.h>
int main(void){
unsigned long long nth = 1000000000000ULL-1;//-1: 1 origin
unsigned long long k[30] = {0, 5};//28:Log10(ULL_MAX)/Log10(5) + 1
unsigned long long sum = k[1];
unsigned long long temp = 4;//4 : 2,4,6,8
int i;
//make table
for(i = 1; i < 30 ; ++i){
if(k[i] == 0){
temp *= 5;//5 : 0,2,4,6,8 , 2digits: 4*5, 3digits: 4*5*5
sum += temp;
k[i] = sum;
}
if(k[i] > nth)
break;
}
while(--i){//The same basically barakmanos
int n = nth / k[i];
printf("%d", n * 2);
nth -= n * k[i];
}
printf("%d\n", nth * 2);
return 0;
}