53 lines
1.9 KiB
Java
53 lines
1.9 KiB
Java
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package Backtracking;
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import java.util.Scanner;
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/*
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* Problem Statement :
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* Find the number of ways that a given integer, N , can be expressed as the sum of the Xth powers of unique, natural numbers.
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* For example, if N=100 and X=3, we have to find all combinations of unique cubes adding up to 100. The only solution is 1^3+2^3+3^3+4^3.
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* Therefore output will be 1.
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*/
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public class PowerSum {
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public static void main(String[] args) {
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Scanner sc = new Scanner(System.in);
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System.out.println("Enter the number and the power");
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int N = sc.nextInt();
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int X = sc.nextInt();
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PowerSum ps = new PowerSum();
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int count = ps.powSum(N,X);
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//printing the answer.
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System.out.println("Number of combinations of different natural number's raised to "+X+" having sum "+N+" are : ");
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System.out.println(count);
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sc.close();
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}
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private int count = 0,sum=0;
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public int powSum(int N, int X) {
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Sum(N,X,1);
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return count;
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}
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//here i is the natural number which will be raised by X and added in sum.
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public void Sum(int N, int X,int i) {
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//if sum is equal to N that is one of our answer and count is increased.
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if(sum == N) {
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count++;
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return;
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}
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//we will be adding next natural number raised to X only if on adding it in sum the result is less than N.
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else if(sum+power(i,X)<=N) {
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sum+=power(i,X);
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Sum(N,X,i+1);
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//backtracking and removing the number added last since no possible combination is there with it.
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sum-=power(i,X);
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}
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if(power(i,X)<N) {
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//calling the sum function with next natural number after backtracking if when it is raised to X is still less than X.
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Sum(N,X,i+1);
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}
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}
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//creating a separate power function so that it can be used again and again when required.
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private int power(int a , int b ){
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return (int)Math.pow(a,b);
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}
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}
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