Med tanke på ett nummer n hitta två par som kan representera antalet som summan av två kuber. Med andra ord hittar två par (a b) och (c d) så att angivet antal n kan uttryckas som
n = a^3 + b^3 = c^3 + d^3
där A B C och D är fyra distinkta siffror.
karta i java
Exempel:
Input: N = 1729 Output: (1 12) and (9 10) Explanation: 1729 = 1^3 + 12^3 = 9^3 + 10^3 Input: N = 4104 Output: (2 16) and (9 15) Explanation: 4104 = 2^3 + 16^3 = 9^3 + 15^3 Input: N = 13832 Output: (2 24) and (18 20) Explanation: 13832 = 2^3 + 24^3 = 18^3 + 20^3
Varje nummer N som uppfyller begränsningen kommer att ha två distinkta par (A B) och (C D) så att A B C och D alla är mindre än n1/3 . Idén är väldigt enkel. För varje distinkt par (x y) bildas av siffror mindre än n1/3 Om deras summa (x3+ och3) är lika med angivet nummer vi lagrar dem i en hashtabell med summan som en nyckel. Om par med summa som är lika med givet nummer visas igen skriver vi helt enkelt ut båda paren.
1) Create an empty hash map say s. 2) cubeRoot = n1/3 3) for (int x = 1; x < cubeRoot; x++) for (int y = x + 1; y <= cubeRoot; y++) int sum = x3 + y3; if (sum != n) continue; if sum exists in s we found two pairs with sum print the pairs else insert pair(x y) in s using sum as key
Nedan är implementeringen av ovanstående idé -
mergesort javaC++
// C++ program to find pairs that can represent // the given number as sum of two cubes #include
// Java program to find pairs that can represent // the given number as sum of two cubes import java.util.*; class GFG { static class pair { int first second; public pair(int first int second) { this.first = first; this.second = second; } } // Function to find pairs that can represent // the given number as sum of two cubes static void findPairs(int n) { // find cube root of n int cubeRoot = (int) Math.pow(n 1.0/3.0); // create an empty map HashMap<Integer pair> s = new HashMap<Integer pair>(); // Consider all pairs such with values less // than cuberoot for (int x = 1; x < cubeRoot; x++) { for (int y = x + 1; y <= cubeRoot; y++) { // find sum of current pair (x y) int sum = x*x*x + y*y*y; // do nothing if sum is not equal to // given number if (sum != n) continue; // if sum is seen before we found two pairs if (s.containsKey(sum)) { System.out.print('(' + s.get(sum).first+ ' ' + s.get(sum).second+ ') and (' + x+ ' ' + y+ ')' +'n'); } else // if sum is seen for the first time s.put(sum new pair(x y)); } } } // Driver code public static void main(String[] args) { int n = 13832; findPairs(n); } } // This code is contributed by PrinciRaj1992
# Python3 program to find pairs # that can represent the given # number as sum of two cubes # Function to find pairs that # can represent the given number # as sum of two cubes def findPairs(n): # Find cube root of n cubeRoot = pow(n 1.0 / 3.0); # Create an empty map s = {} # Consider all pairs such with # values less than cuberoot for x in range(int(cubeRoot)): for y in range(x + 1 int(cubeRoot) + 1): # Find sum of current pair (x y) sum = x * x * x + y * y * y; # Do nothing if sum is not equal to # given number if (sum != n): continue; # If sum is seen before we # found two pairs if sum in s.keys(): print('(' + str(s[sum][0]) + ' ' + str(s[sum][1]) + ') and (' + str(x) + ' ' + str(y) + ')' + 'n') else: # If sum is seen for the first time s[sum] = [x y] # Driver code if __name__=='__main__': n = 13832 findPairs(n) # This code is contributed by rutvik_56
// C# program to find pairs that can represent // the given number as sum of two cubes using System; using System.Collections.Generic; class GFG { class pair { public int first second; public pair(int first int second) { this.first = first; this.second = second; } } // Function to find pairs that can represent // the given number as sum of two cubes static void findPairs(int n) { // find cube root of n int cubeRoot = (int) Math.Pow(n 1.0/3.0); // create an empty map Dictionary<int pair> s = new Dictionary<int pair>(); // Consider all pairs such with values less // than cuberoot for (int x = 1; x < cubeRoot; x++) { for (int y = x + 1; y <= cubeRoot; y++) { // find sum of current pair (x y) int sum = x*x*x + y*y*y; // do nothing if sum is not equal to // given number if (sum != n) continue; // if sum is seen before we found two pairs if (s.ContainsKey(sum)) { Console.Write('(' + s[sum].first+ ' ' + s[sum].second+ ') and (' + x+ ' ' + y+ ')' +'n'); } else // if sum is seen for the first time s.Add(sum new pair(x y)); } } } // Driver code public static void Main(String[] args) { int n = 13832; findPairs(n); } } // This code is contributed by PrinciRaj1992
// JavaScript program to find pairs that can represent // the given number as sum of two cubes // Function to find pairs that can represent // the given number as sum of two cubes function findPairs(n){ // find cube root of n let cubeRoot = Math.floor(Math.pow(n 1/3)); // create an empty map let s = new Map(); // Consider all pairs such with values less // than cuberoot for (let x = 1; x < cubeRoot; x++){ for (let y = x + 1; y <= cubeRoot; y++){ // find sum of current pair (x y) let sum = x*x*x + y*y*y; // do nothing if sum is not equal to // given number if (sum != n){ continue; } // if sum is seen before we found two pairs if (s.has(sum)){ console.log('(' s.get(sum)[0] '' s.get(sum)[1] ') and ('x'' y')'); } else{ // if sum is seen for the first time s.set(sum [x y]); } } } } // Driver function { let n = 13832; findPairs(n); } // The code is contributed by Gautam goel (gautamgoel962)
Produktion:
(2 24) and (18 20)
Tidskomplexitet av ovanstående lösning är o (n2/3) vilket är mycket mindre än O (n).
Kan vi lösa ovanstående problem i o (n1/3) tid? Vi kommer att diskutera det i nästa inlägg.