Use do–while loop and set the condition to while(L != ‘1’ && L != ‘2’ && L != ‘3’);: do { L = getch(); switch (L) { case ‘1’ : system(“cls”); printf(“111111111111111”); break; case ‘2’ : system(“cls”); printf(“222222222222222”); break; case ‘3’ : system(“cls”); printf(“33333333”); break; default : sleep(0); } } while(L != ‘1’ && L … Read more
Because that is not a valid expression. Change this: while (player_bet >= 0 && <= end_money) To: while (player_bet >= 0 && player_bet <= end_money) Translation: while player-bet is 0 or bigger, and also while player-bet is end-money or smaller. Your original expression is roughly: while player_bet is zero or bigger and also while (something … Read more
for (int i=12; i>=1; i–) { String month; if (i<10) month= “0” + i; else month= “” + i; //You might want to do something with the variable “month” }
A for loop would be a good choice in this case. To give you a head start: #Take the smallest and largest numbers in input for(float i = smallest ; i <= largest ; i = i+6 ){ #do your calculations #Print out the number , endline }
If the number input isn’t one that you check (1 to 5) then you hit: else if (a == 1) { std::cout << “\n”; return 1; } which will enter the if (because a is 1), print a new line and return from main ending your run.
If you want a faster solution you can use itertools.repeat: from itertools import repeat: list(zip(repeat(0), range(101))) benchmark: %timeit [(0, x) for x in range(101)] # 3.64 µs ± 19.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each) %timeit list(zip(repeat(0), range(101))) # 2.81 µs ± 35.3 ns per loop (mean ± … Read more
count starts at 0 and number 2 is added to it at each iteration. The loop stops when count is equal or higher than 10. Therefore the values of count will be : 0, 2, 4, 6, 8, 10 These are all even numbers so count % 2 == 0 will be true. result starts … Read more
Python understands 1.2e34, as a float, but you can cast it to an int. int(1.2e34). If you want to loop between 1 and n inclusive, you would normally use range(1, n+1). Thus, in Python 3: for i in range(1, int(1.2e34)+1): print(i) # or do whatever you want — As FHTMitchell pointed out, in Python 2, … Read more
You need to have var myArray = [] and then print the array.
This seems like it was purposefully un-optimized. I would assume that the time constraints are coming from modding a variable 100,000,000,000 times. But wait, that is not all. Not only do you mod a variable that many times, but when a variable modded to 0, you iterate another 11 times over each char in “Hello … Read more