Ja ir pozitīvs skaitlis N, mums jāsasniedz 1 minimālajā soļu skaitā, kur solis tiek definēts kā N pārveidošana par (N-1) vai N pārvēršana par vienu no lielākajiem dalītājiem.
Formāli, ja esam pie N, tad 1 solī varam sasniegt (N - 1) vai, ja N = u*v, tad varam sasniegt max(u v), kur u > 1 un v > 1.
Piemēri:
Input : N = 17 Output : 4 We can reach to 1 in 4 steps as shown below 17 -> 16(from 17 - 1) -> 4(from 4 * 4) -> 2(from 2 * 2) -> 1(from 2 - 1) Input : N = 50 Output : 5 We can reach to 1 in 5 steps as shown below 50 -> 10(from 5 * 10) -> 5(from 2 * 5) -> 4(from 5 - 1) -> 2(from 2 *2) -> 1(from 2 - 1)
Mēs varam atrisināt šo problēmu, izmantojot pirmo platuma meklēšanu, jo tā darbojas līmenī pēc līmeņa, tāpēc mēs sasniegsim līdz 1 minimālajā soļu skaitā, kur nākamajā N līmenī ir (N - 1) un lielāki N pareizie faktori.
Pilnīga BFS procedūra būs šāda Vispirms mēs datu rindā ievietosim N ar soļiem 0, tad katrā līmenī mēs ievietosim to nākamā līmeņa elementus ar 1 soli vairāk nekā tā iepriekšējā līmeņa elementi. Tādā veidā, kad 1 tiks izmests no rindas, tajā būs minimālais soļu skaits, kas būs mūsu gala rezultāts.
Zemāk esošajā kodā tiek izmantota “datu” tipa struktūras rinda, kas saglabā vērtību un soļus no N tajā, tiek izmantota cita vesela skaitļa tipa kopa, lai pasargātu mūs no viena un tā paša elementa nospiešanas vairāk nekā vienu reizi, kas var izraisīt bezgalīgu cilpu. Tāpēc katrā solī mēs nospiežam vērtību komplektā pēc tās ievietošanas rindā, lai vērtība netiktu apmeklēta vairāk kā vienu reizi.
Lai labāk izprastu, lūdzu, skatiet tālāk norādīto kodu
C++// C++ program to get minimum step to reach 1 // under given constraints #include
// Java program to get minimum step to reach 1 // under given constraints import java.util.*; class GFG { // structure represent one node in queue static class data { int val; int steps; public data(int val int steps) { this.val = val; this.steps = steps; } }; // method returns minimum step to reach one static int minStepToReachOne(int N) { Queue<data> q = new LinkedList<>(); q.add(new data(N 0)); // set is used to visit numbers so that they // won't be pushed in queue again HashSet<Integer> st = new HashSet<Integer>(); // loop until we reach to 1 while (!q.isEmpty()) { data t = q.peek(); q.remove(); // if current data value is 1 return its // steps from N if (t.val == 1) return t.steps; // check curr - 1 only if it not visited yet if (!st.contains(t.val - 1)) { q.add(new data(t.val - 1 t.steps + 1)); st.add(t.val - 1); } // loop from 2 to Math.sqrt(value) for its divisors for (int i = 2; i*i <= t.val; i++) { // check divisor only if it is not visited yet // if i is divisor of val then val / i will // be its bigger divisor if (t.val % i == 0 && !st.contains(t.val / i) ) { q.add(new data(t.val / i t.steps + 1)); st.add(t.val / i); } } } return -1; } // Driver code public static void main(String[] args) { int N = 17; System.out.print(minStepToReachOne(N) +'n'); } } // This code is contributed by 29AjayKumar
# Python3 program to get minimum step # to reach 1 under given constraints # Structure represent one node in queue class data: def __init__(self val steps): self.val = val self.steps = steps # Method returns minimum step to reach one def minStepToReachOne(N): q = [] q.append(data(N 0)) # Set is used to visit numbers # so that they won't be pushed # in queue again st = set() # Loop until we reach to 1 while (len(q)): t = q.pop(0) # If current data value is 1 # return its steps from N if (t.val == 1): return t.steps # Check curr - 1 only if # it not visited yet if not (t.val - 1) in st: q.append(data(t.val - 1 t.steps + 1)) st.add(t.val - 1) # Loop from 2 to Math.sqrt(value) # for its divisors for i in range(2 int((t.val) ** 0.5) + 1): # Check divisor only if it is not # visited yet if i is divisor of val # then val / i will be its bigger divisor if (t.val % i == 0 and (t.val / i) not in st): q.append(data(t.val / i t.steps + 1)) st.add(t.val / i) return -1 # Driver code N = 17 print(minStepToReachOne(N)) # This code is contributed by phasing17
// C# program to get minimum step to reach 1 // under given constraints using System; using System.Collections.Generic; class GFG { // structure represent one node in queue class data { public int val; public int steps; public data(int val int steps) { this.val = val; this.steps = steps; } }; // method returns minimum step to reach one static int minStepToReachOne(int N) { Queue<data> q = new Queue<data>(); q.Enqueue(new data(N 0)); // set is used to visit numbers so that they // won't be pushed in queue again HashSet<int> st = new HashSet<int>(); // loop until we reach to 1 while (q.Count != 0) { data t = q.Peek(); q.Dequeue(); // if current data value is 1 return its // steps from N if (t.val == 1) return t.steps; // check curr - 1 only if it not visited yet if (!st.Contains(t.val - 1)) { q.Enqueue(new data(t.val - 1 t.steps + 1)); st.Add(t.val - 1); } // loop from 2 to Math.Sqrt(value) for its divisors for (int i = 2; i*i <= t.val; i++) { // check divisor only if it is not visited yet // if i is divisor of val then val / i will // be its bigger divisor if (t.val % i == 0 && !st.Contains(t.val / i) ) { q.Enqueue(new data(t.val / i t.steps + 1)); st.Add(t.val / i); } } } return -1; } // Driver code public static void Main(String[] args) { int N = 17; Console.Write(minStepToReachOne(N) +'n'); } } // This code is contributed by 29AjayKumar
<script> // Javascript program to get minimum step // to reach 1 under given constraints // Structure represent one node in queue class data { constructor(val steps) { this.val = val; this.steps = steps; } } // Method returns minimum step to reach one function minStepToReachOne(N) { let q = []; q.push(new data(N 0)); // Set is used to visit numbers // so that they won't be pushed // in queue again let st = new Set(); // Loop until we reach to 1 while (q.length != 0) { let t = q.shift(); // If current data value is 1 // return its steps from N if (t.val == 1) return t.steps; // Check curr - 1 only if // it not visited yet if (!st.has(t.val - 1)) { q.push(new data(t.val - 1 t.steps + 1)); st.add(t.val - 1); } // Loop from 2 to Math.sqrt(value) // for its divisors for(let i = 2; i*i <= t.val; i++) { // Check divisor only if it is not // visited yet if i is divisor of val // then val / i will be its bigger divisor if (t.val % i == 0 && !st.has(t.val / i)) { q.push(new data(t.val / i t.steps + 1)); st.add(t.val / i); } } } return -1; } // Driver code let N = 17; document.write(minStepToReachOne(N) + '
'); // This code is contributed by rag2127 </script>
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