(5/5)

Just fill in missing codes

For the integers x in the range x [ LB,UB ], find those x that can be expressed as the sum of one or more distinct squares and show the squares; count those x that cannot be expressed as the sum of one or more distinct squares; and distinguish each x as either prime or composite. For example, the results for integers in the range [ 1,30 ] are shown below in the “Sample Program Dialog”.

Notes

• 1 is neither prime nor composite, but can be expressed as the sum of exactly one distinct square, namely, 1 = 1^2

• 2 and 3 are prime numbers that cannot be expressed as the sum of one or more distinct squares (however 5, 13, 17, and 29 are prime numbers that can be expressed as the sum of one or more distinct squares)

• 6, 8, 12, 15, 18, 22, 24, 27, and 28 are composite numbers that cannot be expressed as the sum of one or more distinct squares

• 25, 26, 29, and 30 are composite numbers that can be expressed as the sum of one or more distinct squares and in more than one way!

Unfortunately, the claim there are exactly 37 integers in [ 1,149 ] that cannot be expressed as the sum of one or more distinct squares is not true! Run your version of Problem48.c to prove that the claim is not true! So, then, how many integers and which integers in the range [ 1,149 ] cannot be expressed as the sum of one or more distinct squares?!

Sample Program Dialog

LB? 1

UB? 30

1 = 1^2

*** 1: 2 (prime)

*** 2: 3 (prime)

4 = 2^2

5 = 1^2 + 2^2 (prime)

*** 3: 6

*** 4: 7 (prime)

*** 5: 8

9 = 3^2

10 = 1^2 + 3^2

*** 6: 11 (prime)

*** 7: 12

13 = 2^2 + 3^2 (prime)

14 = 1^2 + 2^2 + 3^2

*** 8: 15

16 = 4^2

17 = 1^2 + 4^2 (prime)

*** 9: 18

***10: 19 (prime)

20 = 2^2 + 4^2

21 = 1^2 + 2^2 + 4^2

***11: 22

***12: 23 (prime)

***13: 24

25 = 5^2

25 = 3^2 + 4^2

26 = 1^2 + 5^2

26 = 1^2 + 3^2 + 4^2

***14: 27

***15: 28

29 = 2^2 + 5^2 (prime)

29 = 2^2 + 3^2 + 4^2

30 = 1^2 + 2^2 + 5^2

30 = 1^2 + 2^2 + 3^2 + 4^2

LB? ^Z

Press any key to continue . . .

Computational thinking questions.

1. Write a for-statement that allows the int-eger x to traverse every int-eger in the int-eger range

x [ LB,UB ].

2. What is the variable count counting?! Hint The maximum value of count is (UB-LB+1).

//-------------------------------------------------------------

// Problem #48

// Problem48.c

//-------------------------------------------------------------

#include <stdio.h>

#include <stdlib.h>

#include <stdbool.h>

#include <math.h>

#include ".\Combinations.h"

//-------------------------------------------------------------

int main()

//-------------------------------------------------------------

{

bool IsPrime(const int x);

int LB,UB;

printf("LB? ");

while ( scanf("%d",&LB) != EOF )

{

int n;

int count;

printf("UB? "); scanf("%d",&UB);

n = (int) (sqrt(UB)+0.5);

count = 0;

for (int x = LB; x <= UB; x++)

{

bool isExpressable = false;

bool isExpression1 = true;

for (int k = 1; k <= n; k++)

{

COMBINATIONS C;

ConstructCombinations(&C,n,k);

FindFirstCombination(&C);

do

{

Student provides missing code to consider all k-combinations, k [ 1,n ], for each x, then

display each of the following datums using formats inferred from the "Sample Program Dialog"

(1) x itself;

(2) x's expression "as the sum of one or more distinct squares" (when applicable);

(3) the count of xs that cannot be expressed "as the sum of one or more distinct squares"

(when applicable); and

(4) x's prime-ness (when applicable).

*Note* The code that satisfies some of the foregoing requirements is provided both above and below!?

FindNextCombination(&C);

} while ( !AtEndOfCombinations(&C) );

DestructCombinations(&C);

}

if ( !isExpressable )

{

count++;

printf("***%2d: %7d",count,x);

if ( IsPrime(x) )

printf(" (prime)");

printf("\n");

}

}

printf("\nLB? ");

}

system("PAUSE");

return( 0 );

}

//-------------------------------------------------------------

bool IsPrime(const int x)

//-------------------------------------------------------------

{

bool r;

if ( x == 1 )

r = false;

else

{

r = true;

for (int i = 2; i <= (int) (sqrt(x)+0.5); i++)

r = r && (x%i != 0);

}

return( r );

}

(5/5)

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