How to add multiple numbers using nested loops?-CodePudding

I am doing an assignment where I must use nested loops in order to add up the squares and cubes of integers from 1 to N (N being whatever the user inputs). For example, if the user input the number 5, the program is supposed to do "1² 2² 3² 4² 5²" and output the sum of those numbers, as well "1³ 2³ 3³ 4³ 5³" and output the sum of those numbers.

However, I am having trouble figuring out how to code it in a way that I receive the proper output? This is what I wrote. Scanners were already added.

int limitNum = input.nextInt();
double squareNums:
double sumofSq = 0;
double cubedNums;
double sumofCubes = 0;


for(int s = 1; s <= limitNum; s  )
    {
        for(int c = 1; c <= limitNum; c  )
           {
     cubedNums = Math.pow(c, 3);
     sumofCubes = sumofCubes   cubedNums;
           }
       squareNums= Math.pow(s, 2);
       sumofSq = sumofSq   squareNums;
    }

But currently, when I run this program, the sum of the squares output correctly, but the sum of the cubes is always some big number. For example if 5 is used, sumofSq would output 55.0, but sumofCubes would output 1125.0.

CodePudding user response：

There is no point using a nested loop as this would result in complexity of O(n²). A single loop would be sufficient and be in complexity class O(n).

public class Application {
    public static void main(String[] args) {
        var squareSum = 0d;
        var cubeSum = 0d;
        var upperBound = 5;
        for(var i = 1; i <= upperBound; i  ){
            squareSum  = Math.pow(i, 2);
            cubeSum  = Math.pow(i, 3);
        }
        System.out.printf("""
                Sum of first n squares: %s
                Sum of first n cubes: %s
                """, (int)squareSum, (int)cubeSum);
    }
}

In fact there is no need to loop at all => constant computation time no matter the size of the input O(1). There is a well known formula which tells you the sum of the first n squares.

n(n 1)(2n 1)
------------ 
     6

Please see this for a proof.

The same holds true for the first n cubes.

n²(n 1)²
-------- 
   4

Please see this for a proof.

The following program will therefore return the same result.

public class Application {
    public static void main(String[] args) {
        var upperBound = 5;
        System.out.printf("""
                Sum of first n squares: %s
                Sum of first n cubes: %s
                """, sumOfFirstNSquares(upperBound), sumOfFirstNCubes(upperBound));
    }

    public static int sumOfFirstNSquares(int n){
        return (n * (n 1) * (2 * n   1)) / 6;
    }

    public static int sumOfFirstNCubes(int n){
        return ((n * n) * (n 1) * (n 1)) / 4;
    }
}

CodePudding user response：

In fact there is no need to loop at all => constant computation time no matter the size of the input O(1). There is a well known formula which tells you the sum of the first n squares.

java