I'm working on c++ and I need help writing a program that prints all numbers between 1 and 200 excluding multiples of 7. I understand how to print the number between 1 and 200, but I don't understand how to print them excluding multiples of 7.
Use the modulo operator % to determine if the number is a multiple of 7.
if number % 7 == 0, then it's a multiple of seven, and then you don't print it.
You can pre calculate the multiples of 7, storing them somewhere.
Then, for every value in the 'for' you print only if the number is not in where you stored the pre calculations.
Related
In my code, legends are running within a loop, and I am trying to show a graph with
0-10%
10-20%
and so on. The problem is when I write this code
legend->AddEntry(gr[i], Form("%d0-%d0 %%",i+0,i+1), "lep");
It shows
00-10%
10-20% etc
So how to not show 00, but 0 in the first line?
A small adaptation of the shown statement should be enough; use:
legend->AddEntry(gr[i], Form("%d-%d %%", i*10 , (i+1)*10), "lep");
Explanation:
Form("%d0-%d0 %%",i+0,i+1) seems to be some kind of string formatting, and i your loop variable which runs from 0 to 9, right? The shown Form statement just appends "0" hard-coded to the single digit in i; instead, you can multiply i by 10, resulting in the actual numbers you want printed; and since 10*0 is still 0, this will be a single digit still; so, replace the previous Form(...) call with Form("%d-%d %%", i*10, (i+1)*10) and you should have the result you want!
In case you're worrying that printing i*10 is "less efficient" than printing i with "0" suffix - don't. The formatting and output of the string is most probably orders of magnitude slower than the multiplication anyway, so any overhead of doing multiple multiplications is negligible.
I know this sounds like a very specific question, but I am making a key generator for a program, and one part of the key has seven digits that need to be a multiple of seven when added together. How do you achieve this?
Generate a random 6-digit number then choose the number for the 7th digit that makes the algorithm work.
See any example of how to use std::rand for generating the random digits.
Once you have generated the 7 numbers for the key. You could than try a %7.
If the result is 0, than it mean that your number generated is a multiple of 7.
If the result is not 0, than you just have to regenerate another key or add 1 to the last number until modulo 7 (%7) equal 0.
How to generate a fix length 14 digit long random number. I am using postman and lodash lib but somehow sometimes it works sometimes it doesn't
pm.globals.set('randomNumer_SO',_.random(1, 100000000000000));
This sometimes generate 13 digit number too. But I always want a 14 digit as else API fails.
When you use _.random(1, 100000000000000) you ask for a number between 1 (length 1), and 100000000000000 (length 15). Most of the time, you'll get numbers with length 14, but you might also get numbers of length 1, 2, 3, etc...
To get a number with fixed length of 14, set the lower and upper bounds to the minimal and maximal numbers with that length:
_.random(10000000000000, 99999999999999)
i have made a simple bit of code to test a number to see if it is a prime number or not but while feeding it large prime numbers to test the speed of the program on the arduino it whould only take a number at length 9-/under-digits i tested my read function and it returns the entire number but the 'BigNumber' wont parse it insted it just says its 0
code:
void Speed(String num)
{
Serial.println("NUM="+num);
BigNumber NUM = num.c_str();//this is where it fails
BigNumber Curr = "1";//start 2 / 'curr++' start of loop
num = "";
... the testing of prime numbers here
the code stops the arduino if i put a 10 digit number in, the output is so
<|S 1234567891
>|NUM=1234567891
and if i put a number with 9 digits it outputs as expected
<S 123456789
>|NUM=123456789
>|123456789 is not a prime number
>|because ist a factor of 3
i have tryed seeing if anyone has had the same problem as me but i cant find it anywhere.
im use an arduino-uno
EDIT: after doing some more testing it now doesnt set the number insted of crashing after testing 'S 1111111111' (10 digits) its output is normal:
<|S 1111111111
>|NUM=1111111111
>|1111111111 is not a prime number
>|because ist a factor of 11
but if i put in 11 digits it parses as 0 ??
<|S 11111111111
>|NUM=11111111111
>|0 cant be a prime number because it doesn't end in 1,3,7,9
bty: i forgot to mention that 'S number_here' S specifys the mothod of finding the result i also have D=DataCrunch it checks all the numbers and L=List witch creates a list of found Prime numbers like a prime number search, and thay work fine exept that DataCrunch (D) has the same problem with parsing the number given.
EDIT2:
this is proof that BigNumber can hold such a large number
https://forum.arduino.cc/index.php?topic=85692.0
in the first post.
so as it turns out after some extensiv research that BigNumber is not fit for very large numbers but another part of the 'BigNumber.h' lib does its bc_num.
bc_num x;
bc_str2num(&x, "9898989898", 10);
String c = "Controll=";
c+=bc_num2str(x);
Serial.println(c);
output
Controll=9898989898
but as you can see this takes a bit more programming to get implamented and so im going to go off and start now bye.
An emirp (prime spelled backwards) is a pime number whose reversal is also prime. Ex. 17 & 71. I have to write a program that displays the first 100 emirps. It has to display 10 numbers per line and align the numbers properly:
2 3 5 7 11 13 17 31 37 71
73 79 97 101 107 113 131 149 151 157.
I have no cue what I am doing and would love if anyone could dump this down for me.
It sounds like there are two general problems:
Finding the emirps.
Formatting the output as required.
Break down your tasks into smaller parts, and then you'll be able to see more clearly how to get the whole task done.
To find the emirps, first write some helper functions:
is_prime() to determine whether a number is prime or not
reverse_digits() to reverse the digits of any number
Combining these two functions, you can imagine a loop that finds all the numbers that are primes both forward and reversed. Your first task is complete when you can simply generate a list of those numbers, printing them to the console one per line.
Next, work out what format you want to use (it looks like a fixed format of some number of character spaces per number is what you need). You know that you have 100 numbers, 10 per line, so working out how to format the numbers should be straightforward.
Break the problem down into simpler sub-problems:
Firstly, you need to check whether a number is prime. This is such a common task that you can easily Google it - or try a naive implementation yourself, which may be better given that this is homework.
Secondly, you need to reverse the digits of a number. I'd suggest you figure out an algorithm for this on a piece of paper first, then implement it in code.
Put the two together - it shouldn't be that hard.
Format the results properly. Printing 10 numbers per line is something you should be able to figure out easily once the rest is done.
Once you have a simple version working you might be able to optimise it in some way.
A straight forward way of checking if a number is prime is by trying all known primes less than it and seeing if it divides evenly into that number.
Example: To find the first couple of primes
Start off with the number 2, it is prime because the only divisors are itself and 1, meaning the only way to multiple two numbers to get 2 is 2 x 1. Likewise for 3.
So that starts us off with two known primes 2 and 3. To check the next number we can check if 4 modulo 2 equals 0. This means when divide 2 into 4 there is no remainder, which means 2 is a factor of 4. Specifically we know 2 x 2 = 4. Thus 4 is not prime.
Moving on to the next number: 5. To check five we try 5 modulo 2 and 5 modulo 3, both of which equals one. So 5 is prime, add it to our list of known primes and then continue on checking the next number. This rather tedious process is great for a computer.
So on and so forth - check the next number by looping through all previous found primes and check if they divide evenly, if all previously found primes don't divide evenly, you have a new prime. Repeat.
You can speed this up by counting by 2's, since all even numbers are divisible by two. Also, another nice trick is you don't have to check any primes greater than the square root of the number, since anything larger would need a smaller prime factor. Cuts your loops in half.
So that is your algorithm to generate a large list of primes.
Collect a good chunk of them in an array, say the first 10,000 or so. And then loop through them, reverse the numbers and see if the result is in your array. If so you have a emirp, continue until you get the first 100 emirps
If the first 10,000 primes don't return 100 emirps. Move on to the next 10,000. Repeat.
For homework, I would use a fairly simplistic isPrime function, pseudo-code along the lines of:
def isPrime (num):
set testDiv1 to 2
while testDiv1 multiplied by testDiv1 is less than or equal to num:
testDiv2 = integer part of (num divided by testDiv1)
if testDiv1 multiplied by testDiv2 is equal to num:
return true
Add 1 to testDiv1
return false
This basically checks whether the number is evenly divisible by any number between 2 and the square root of the number, a primitive primality check. The reson you stop at the square root is because you would have already found a match below it if there was one above it.
For example 100 is 2 times 50, 4 times 25, 5 time 20 and 10 times 10. The next one after that would be 20 times 5 but you don't need to check 20 since it would have been found when you checked 5. Any positive number can be expressed as a product of two other positive numbers, one below the square root and one above (other than the exact square root case of course).
The next tricky bit is the reversal of digits. C has some nice features which will make this easier for you, the pseudo-code is basically:
def reverseDigits (num):
set newNum to zero
while num is not equal to zero:
multiply newnum by ten
add (num modulo ten) to newnum
set num to the integer part of (num divided by ten)
return newNum
In C, you can use int() for integer parts and % for the modulo operator (what's left over when you divide something by something else - like 47 % 10 is 7, 9 % 4 is 1, 1000 % 10 is 0 and so on).
The isEmirp will be a fairly simplistic:
def isEmirp (num):
if not isPrime (num):
return false
num2 = reverseDigits (num)
if not isPrime (num2):
return false
return true
Then at the top level, your code will look something like:
def mainProg:
create array of twenty emirps
set currEmirp to zero
set tryNum to two
while currEmirp is less than twenty
if isEmirp (tryNum):
put tryNum into emirps array at position currEmirp
add 1 to currEmirp
for currEmirp ranging from 0 to 9:
print formatted emirps array at position currEmirp
print new line
for currEmirp ranging from 10 to 19:
print formatted emirps array at position currEmirp
print new line
Right, you should be able to get some usable code out of that, I hope. If you have any questions of the translation, leave a comment and I'll provide pointers for you, rather than solving it or doing the actual work.
You'll learn a great deal more if you try yourself, even if you have a lot of trouble initially.