/*This code is a general version of sum as cubes program . Problem : Write a program to display the number as the sum of npowers with minimum number of operations. Input the Number first , followed by the exponent in which sum should be displayed . And lastly enter true or false if you wish to know all possible summations. for example : Input : 25 2 false Output : Number entered : 25 The mininum number of operations are 1 5^2 Caution : Don't enter large values . Can result in TLE or MLE. */ import java.util.*; import java.util.stream.Collectors; public class Sum_as_n_powers { public static int minsum , b=0; public static List> npowers = new ArrayList>(); public static void main(String[] args) { try{ Scanner sc= new Scanner(System.in); int num=sc.nextInt() ,n=sc.nextInt(), k=0 ; boolean bool = sc.nextBoolean(); System.out.println("Number entered : "+num); minsumnpowers(num,n,bool) ; System.out.println("\nThe mininum number of operations are " + minsum+"\n"); for(int i=0;i=2 && n==3) System.out.println("This is a Ramanujan Hardy Number"); } catch (Exception e){ System.out.println("Please enter a number"); } } public static void minsumnpowers(int num,int ex , boolean bool){ List> arr = new ArrayList<>(); int a=0; minsum=num; sum(num,ex,new ArrayList<>(),0); npowers=npowers.stream().distinct().collect(Collectors.toList()); // System.out.println(npowers.toString()); if(bool) {System.out.println("All Possible operations :"); for(int i=0;i npower,int n){ int limit = nroot(num,ex) , temp = num ,templimit = limit , i = limit ; List npower1= new ArrayList<>(npower); while ( i > 0){ temp-=Math.pow(templimit,ex); npower.add(n++,templimit); if (temp==0) { // System.out.println(npower.toString()); if(minsum>=n) { minsum=n;} Collections.sort(npower); npowers.add(b++,npower); npower = new ArrayList<>(); npower.addAll(npower1); i--; temp=num; templimit=i; n=npower.size(); continue ; } templimit=(int)nroot(temp,ex); List npower2= new ArrayList<>(npower); if(temp>0){ sum(temp,ex,npower2,n);} else return ; } } public static int nroot(int num , int n){ double root =Math.pow(Math.E,(Math.log(num)/n)); // System.out.println(root); root+=0.000000000001; return (int)root; } }

Sum as power of N