Розбір

Let $f(n)$ represent the minimum number of operations required to transform the number $n$ into $1$. For example,

• $f(1)=0$, as we already have the number $1$;

• $f(2)=1$, perform operation $2\to 1$;

• $f(5)=3$, perform operations $5\to 4\to 2\to 1$;

• $f(10)=3$, perform operations $10\to 9\to 3\to 1$;

In the case of $n=10$, it is more advantageous to subtract $1$ first rather than using the greedy approach and dividing by $2$.

Let’s consider the process of computing the function $f(n)$.

• If we divide the number $n$ by $3$ (assuming $n$ is divisible by $3$), then

  $$f(n)=f(n/3)+1$$

• If we divide the number $n$ by $2$ (assuming $n$ is divisible by $2$), then

  $$f(n)=f(n/2)+1$$

• If we subtract $1$ from the number $n$, then

  $$f(n)=f(n-1)+1$$

From the number $n$, we can obtain one of three numbers: $n/3,n/2$, or $n-1$. The number of operations required to reduce each of these numbers to $1$, is equal to $f(n/3),f(n/2)$, and $f(i-1)$ respectively. Since we are interested in the minimum number of operations, we have the following relationship:

In the case, if $n$ is not divisible by $2$ (or by $3$), then the corresponding element ($f(n/2)$ or $f(n/3)$) is absent in the $min$ function. For example, for $n=8$, we have:

Similarly, for $n=7$, we get:

The values of the function $f(n)$ will be stored in the cells of the array $d[MAX]$, where $MAX=10^6+1$. Fill the cells of the array $d$ from $1$ to $10^6$ according to the given recurrence relation. For example, the following table shows the values of $d[i]$ for $1\le i\le 11$:

For example,

In other words, it is more efficient to subtract $1$ from $10$, rather than divide it by $2$.

Exercise

Find the values of $d[i]$ for the following values of $i$:

Algorithm realization

Declare an array $d$, where the $i$-th cell contains the minimum number of operations required to transform the number $i$ into $1$.

int d[1000001];

Fill the cells of array $d$ according to the provided formulas.

d[1] = 0;
for(i = 2; i <= 1000000; i++)
{
  // d[i] = min(d[i/3],d[i/2],d[i-1]) + 1
  d[i] = d[i-1];
  if (i % 2 == 0 && d[i/2] < d[i]) d[i] = d[i/2];
  if (i % 3 == 0 && d[i/3] < d[i]) d[i] = d[i/3];
  d[i]++;
}

For each input value $n$, print the answer $d[n]$.

while(scanf("%d",&n) == 1)
  printf("%d\n",d[n]);

Algorithm realization — recursion

#include <stdio.h>
#include <string.h>
#define INF 2000000000

int i, res, n;
int dp[1000001];

int min(int a, int b, int c)
{
  int res = a;
  if (b < res) res = b;
  if (c < res) res = c;
  return res;
}

int f(int n)
{
  if (n == 1) return 0;
  if (dp[n] != -1) return dp[n];
  int a = f(n - 1);
  int b = (n % 2 == 0) ? f(n / 2) : INF;
  int c = (n % 3 == 0) ? f(n / 3) : INF;
  return dp[n] = min(a,b,c) + 1;
}

int main(void)
{
  memset(dp, -1, sizeof(dp));
  while (scanf("%d", &n) == 1)
  {
    res = f(n);
    printf("%d\n", res);
  }
  return 0;
}

Java realization

import java.util.*;

public class Main
{
  public static int MAX = 1000001;    
  static int d[] = new int[MAX];

  public static void main(String[] args)
  {
    Scanner con = new Scanner(System.in);
    d[1] = 0;
    for(int i = 2; i < MAX; i++)
    {
      d[i] = d[i-1] + 1;
      if (i % 2 == 0) d[i] = Math.min(d[i], d[i/2] + 1);
      if (i % 3 == 0) d[i] = Math.min(d[i], d[i/3] + 1);
   }

    while(con.hasNext())
    {
      int n = con.nextInt();
      System.out.println(d[n]);
    }
  }
}

Python realization

Create a list $d$, where the $i$-th cell contains the minimum number of operations required to transform the number $i$ into $1$.

d = [0] * 1000001

Fill the cells of array $d$ according to the provided formulas.

d[1] = 0
for i in range(2, 1000001):
  d[i] = d[i - 1]
  if i % 2 == 0 and d[i // 2] < d[i]:
    d[i] = d[i // 2]
  if i % 3 == 0 and d[i // 3] < d[i]:
    d[i] = d[i // 3]
  d[i] += 1

For each input value $n$, print the answer $d[n]$.

while True:
  try:
    n = int(input())
    print(d[n])
  except EOFError:
    break