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I wrote some program in C language. I have to calculate sum of squares. My book suggested me to use loops but I chose more efficient way by using formula:

$$1^2+2^2+\cdots+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$$

Original code from listing 5.13:

#include <stdio.h>

int main(void) {
    int i, sum;

    i = 0;
    sum = 0;
    while (i++ < 20)
        sum = sum + i;
    printf("Sum = %d\n", sum);
    return 0;
}

Finally, this is my code:

#include <stdio.h>

int main(void) {
    int range;
    printf("Enter the range of the calculation:\n");
    scanf("%i", &range);
    printf("Sum of squares = %d\n", (((range * (range + 1)) * (2 * range + 1)) / 6));
    return 0;
}

What do you think about this solution? Is it more efficient than loops?

I wrote some program in C language. I have to calculate sum of squares. My book suggested me to use loops but I chose more efficient way by using formula:

$$1^2+2^2+\cdots+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$$

Original code from listing 5.13:

#include <stdio.h>

int main(void) {
    int licznik, suma;

    licznik = 0;
    suma = 0;
    while (licznik++ < 20)
        suma = suma + licznik;
    printf("suma = %d\n", suma);
    return 0;
}

Finally, this is my code:

#include <stdio.h>

int main(void) {
    int zakres;
    printf("Podaj zakres przeprowadzanych obliczen:\n");
    scanf("%i", &zakres);
    printf("suma kwadratow = %d\n", (((zakres * (zakres + 1)) * (2 * zakres + 1)) / 6));
    return 0;
}

What do you think about this solution? Is it more efficient than loops?

I wrote some program in C language. I have to calculate sum of squares. My book suggested me to use loops but I chose more efficient way by using formula:

$$1^2+2^2+\cdots+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$$

Original code from listing 5.13:

#include <stdio.h>

int main(void) {
    int licznik, suma;

    licznik = 0;
    suma = 0;
    while (licznik++ < 20)
        suma = suma + licznik;
    printf("suma = %d\n", suma);
    return 0;
}

Finally, this is my code:

#include <stdio.h>

int main(void) {
    int zakres;
    printf("Podaj zakres przeprowadzanych obliczen:\n");
    scanf("%i", &zakres);
    printf("suma kwadratow = %d\n", (((zakres * (zakres + 1)) * (2 * zakres + 1)) / 6));
    return 0;
}

What do you think about this solution? Is it more efficient than loops?

I wrote some program in C language. I have to calculate sum of squares. My book suggested me to use loops but I chose more efficient way by using formula:

$$1^2+2^2+\cdots+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$$

Original code from listing 5.13:

#include <stdio.h>

int main(void) {
    int i, sum;

    i = 0;
    sum = 0;
    while (i++ < 20)
        sum = sum + i;
    printf("Sum = %d\n", sum);
    return 0;
}

Finally, this is my code:

#include <stdio.h>

int main(void) {
    int range;
    printf("Enter the range of the calculation:\n");
    scanf("%i", &range);
    printf("Sum of squares = %d\n", (((range * (range + 1)) * (2 * range + 1)) / 6));
    return 0;
}

What do you think about this solution? Is it more efficient than loops?

I wrote some program in C language. I have to calculate sum of squares. My book suggested me to use loops but I chose more efficient way by using formula:

$$1^2+2^2+\cdots+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$$

Original code from listing 5.13:

#include <stdio.h>

int main(void) {
    int licznik, suma;

    licznik = 0;
    suma = 0;
    while (licznik++ < 20)
        suma = suma + licznik;
    printf("suma = %d\n", suma);
    return 0;
}

Finally, this is my code:

#include <stdio.h>

int main(void) {
    int zakres;
    printf("Podaj zakres przeprowadzanych obliczen:\n");
    scanf("%i", &zakres);
    printf("suma kwadratow = %d\n", (((zakres * (zakres + 1)) * (2 * zakres + 1)) / 6));
    return 0;
}

What do you think about this solution? Is it more efficient than loops?

I wrote some program in C language. I have to calculate sum of squares. My book suggested me to use loops but I chose more efficient way by using formula:

$$1^2+2^2+\cdots+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$$

Original code from listing 5.13:

#include <stdio.h>

int main(void) {
    int licznik, suma;

    licznik = 0;
    suma = 0;
    while (licznik++ < 20)
        suma = suma + licznik;
    printf("suma = %d\n", suma);
    return 0;
}

Finally, this is my code:

#include <stdio.h>

int main(void) {
    int zakres;
    printf("Podaj zakres przeprowadzanych obliczen:\n");
    scanf("%i", &zakres);
    printf("suma kwadratow = %d\n", (((zakres * (zakres + 1)) * (2 * zakres + 1)) / 6));
    return 0;
}

What do you think about this solution? Is it more efficient than loops?