Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Find Cube Pairs - (A n^(2/3) Solution) in C++
A number is given. We have to find two pairs, that can represent the number as sum of two cubes. So we have to find two pairs (a, b) and (c, d) such that the given number n can be expressed as n = a3 + b3 = c3 + d3
The idea is simple. Here every number a, b, c and d are all less than n1/3. For every distinct pair (x, y) formed by number less than n1/3, if their sum (x3 + y3) is equal to the given number, we store them into hash table with the sum value as key, then if the same sum comes again, them simply print each pair
Algorithm
getPairs(n): begin cube_root := cube root of n map as int type key and pair type value for i in range 1 to cube_root, do for j in range i + 1 to cube_root, do sum = i3 + j3 if sum is not same as n, then skip next part, go to second iteration if sum is present into map, then print pair, and (i, j), else insert (i,j) with corresponding sum into the map done done end
Example
#include <iostream>
#include <cmath>
#include <map>
using namespace std;
int getPairs(int n){
int cube_root = pow(n, 1.0/3.0);
map<int, pair<int, int> > my_map;
for(int i = 1; i<cube_root; i++){
for(int j = i + 1; j<= cube_root; j++){
int sum = i*i*i + j*j*j;
if(sum != n)
continue;
if(my_map.find(sum) != my_map.end()){
cout << "(" << my_map[sum].first << ", " << my_map[sum].second << ") and (" << i << ", " << j << ")" << endl;
}else{
my_map[sum] = make_pair(i, j);
}
}
}
}
int main() {
int n = 13832;
getPairs(n);
}Output
(2, 24) and (18, 20)
Updated on: 25-Sep-2019
67 Views
Advertisements