Can someone help me analyze a question?

Hi

Hi

Anyone Can Help Me please, I have a school assignment Making a calculator with function commands in C code, the calculator calculates the perimeter of a square, rectangle and circle Then I use the switc case to siwtch the calculator menu, Is the switch case included in the function command, or am I doing it wrong?

I think using switch-case is fine in your case. It's not related to functions but it is very basic part of the language, like if.

if with a function, what command do I use?

If you need a function, you can create some and put some of your functionality to be executed in these functions https://www.tutorialspoint.com/cprogramming/c_functions.htm

Hi! What should I do to avoid time limit? #include <stdio.h> long long int number; long long int rax = 2; int main() { scanf("%lld", &number); if (number != 1) { while (number % rax != 0) { rax++; } } else { rax = 1; } printf("%lld", rax); return 0; } Print the smallest divisor of the number n, other than 1.

Why is one variable called "rax"?

rax = score

Why is one variable called "rax"?

so, do u know how I can speed up the process of program?)

Execution speed?

add a if statement to exit the code after testing 50% of the numbers

Try to replace all if's/loops with a bitwise calculation

so, do u know how I can speed up the process of program?)

Hi! What should I do to avoid time limit? #include <stdio.h> long long int number; long long int rax = 2; int main() { scanf("%lld", &number); if (number != 1) { while (number % rax != 0) { rax++; } } else { rax = 1; } printf("%lld", rax); return 0; } Print the smallest divisor of the number n, other than 1.

You can do something like Eratosthene sieve. Array of bools of size - number / 2 representing if given index is a divisor, then for-loop from 2 to number /2, if current index not a divisor, set in loop all multiple values in bool array to false not to check them in future. It's purely algorithmic task, so don't listen to "multithreading" advices

Why is one variable called "rax"?

ukrainian

Why not a even an excellent algorithm can be much faster if parallelized with threads and SIMD that's why there are concurrent and vectorized sort algorithms

i am puzzled on why you would use multi threading for such a simple operation

i am puzzled on why you would use multi threading for such a simple operation

i would of simply %2 != 0 then kicked of a for loop starting at 3 and incrementing by 2 but only checking at 50% of the numbers value rax < (number /2) so if user entered 100 only numbers up to 50 would be tested.

the smallest divisor is never greater than the sqrt of the number

Parallelized bad algo still worse then good algo in 1 thread. And i doubt if this c-like lab work dedicated to teach multithreading. It's overkill for her current knowledge

Hi! What should I do to avoid time limit? #include <stdio.h> long long int number; long long int rax = 2; int main() { scanf("%lld", &number); if (number != 1) { while (number % rax != 0) { rax++; } } else { rax = 1; } printf("%lld", rax); return 0; } Print the smallest divisor of the number n, other than 1.

Hi! What should I do to avoid time limit? #include <stdio.h> long long int number; long long int rax = 2; int main() { scanf("%lld", &number); if (number != 1) { while (number % rax != 0) { rax++; } } else { rax = 1; } printf("%lld", rax); return 0; } Print the smallest divisor of the number n, other than 1.

At very least you can run while upto sqrt (n). No need to check for number above anyway

And you can check that number is odd and make each starting from 3 like rax +=2, you don't need other odd numbers

@Marisichka — Probably she’s finding time-load to find those numbers which are self divisible! Using smallest divisor method with “rax”😄 — #include <stdio.h> #include <time.h> long long int num =0; long long int rax =2; int main() { clock_t tx; scanf("%lld", &num); if(num <2) return 1; for(int i=2; i<=num; i++) { rax=2; tx = clock(); while ((i % rax) != 0) rax++; tx = clock() - tx; if (i==rax) printf("rax[%5.d]:%2.lld |timeTaken:%5.5f\n", i, rax, ((float)tx/CLOCKS_PER_SEC)); } return 0; }

and which problem have you solved here? this is still far from optimal

and which problem have you solved here? this is still far from optimal

@Marisichka — Probably she’s finding time-load to find those numbers which are self divisible! Using smallest divisor method with “rax”😄 — #include <stdio.h> #include <time.h> long long int num =0; long long int rax =2; int main() { clock_t tx; scanf("%lld", &num); if(num <2) return 1; for(int i=2; i<=num; i++) { rax=2; tx = clock(); while ((i % rax) != 0) rax++; tx = clock() - tx; if (i==rax) printf("rax[%5.d]:%2.lld |timeTaken:%5.5f\n", i, rax, ((float)tx/CLOCKS_PER_SEC)); } return 0; }

write a programe that addition two of matrices transpose matrix

write a programe that addition two of matrices transpose matrix

What? 🤔

I have no idea what you're talking about.

write a programe that addition two of matrices transpose matrix

do this programe please

No one is going to write you full program, and even any help is problematic because the question is unclear

If you comment //if(i==rax) This will display the number of iterations “rax” has performed; though I added the for-loop instead of single/single entry!; instead take number for loop and check time limit. She didn’t clear anything on “time limit” question?

From my understanding @Marisichka wanted to calculate minimum using better algorithm because this was timing out.

As original single while without ifs & breaks are best optimized; no improvements needed, unless parallels/openmp required on heavy number crunching! I will leave this to her to answer; as I don’t understand her “time-limit” point correctly; so I played with the program😄

Calculating up to 0.5* sqrt(n) values instead of up to n numbers is even more efficient without any parallelism or vectorisation

Well you are correct to your point & perspectives, but what’s “time-limit” related to?? No clear!🙃

e.g. test with 2147483647, how long does your algorithm take to get the result? more than 10 seconds? that's too slow

1. Can you put the code which is faster! 2. Is that high prime number?

you can get the result for a 15 digit faster so taking 18s for 2147483647 is way to long