Problem:
Score : 300 points

Problem Statement

Takahashi is standing on a multiplication table with infinitely many rows and columns.

The square (i,j) contains the integer i \times j. Initially, Takahashi is standing at (1,1).

In one move, he can move from (i,j) to either (i+1,j) or (i,j+1).

Given an integer N, find the minimum number of moves needed to reach a square that contains N.

Constraints

2 \leq N \leq 10^{12}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the minimum number of moves needed to reach a square that contains the integer N.

Sample Input 1

10

Sample Output 1

5

(2,5) can be reached in five moves. We cannot reach a square that contains 10 in less than five moves.

Sample Input 2

50

Sample Output 2

13

(5, 10) can be reached in 13 moves.

Sample Input 3

10000000019

Sample Output 3

10000000018

Both input and output may be enormous.

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;

namespace Algorithm
{
    class Program
    {
        static List<HashSet<long>> memo;
        static long[] prime;
        static long min;
        static long minA;
        static long minB;
        static void Main(string[] args)
        {
            var N = long.Parse(Console.ReadLine());

            prime = PrimeFactors(N).ToArray();

            min = long.MaxValue;
            minA = 0;
            minB = 0;
            memo = new List<HashSet<long>>();
            Dfs(0, 1, 1);

            Console.WriteLine(minA + minB - 2);

        }

        static void Dfs(int n, long total1, long total2)
        {
           if (n <= memo.Count - 1) {
              if (memo[n].Contains(total1)) return;
              else memo[n].Add(total1);
           }
           else
               memo.Add(new HashSet<long>() {total1});

           if (n == prime.Length) {
              var t = Math.Abs(total1 - total2);
              if (min > t) {
                 min = t;
                 minA = total1;
                 minB = total2; 
              }
              return;
           }
           
           Dfs(n + 1, total1 * prime[n], total2);
           Dfs(n + 1, total1, total2 * prime[n]);
        }
        
        static IEnumerable<long > PrimeFactors(long n)
        {
            long i = 2;
            long tmp = n;

            while (i * i <= n)
            {
                if (tmp % i == 0)
                {
                    tmp /= i;
                    yield return i;
                }
                else
                {
                    i++;
                }
            }
            if (tmp != 1) yield return tmp;
        }
    }
}