So in a for loop, when declaring the temp variable “i” you need to declare it as int i = 1; instead of “i = 1” I assume you already have a value for the variable d but in this case, I just used d = 123. int d = 123; for (int i = … Read more
You need to keep printing space in inner loop and once it come out from the inner loop print * and/or # accordingly. Scanner in = new Scanner(System.in); int x=0; System.out.println(“Enter number: “); x = in.nextInt(); for(int i=1;i<=x;i++){ for(int j=1;j<=i;j++){ System.out.print(” “); } if(i%2==1) { System.out.println(“*”); }else { System.out.println(“#”); } } >>>Demo Link<<< 0 solved … Read more
① You are nesting lists although you just want a flat list (of functions). See the answer of @BartoszKP on that. ② You want to create functions based on local variables. You can do this using lambdas, as @Harsh is proposing, or you can do it using defaulted variables: def F_analytic(k, m, c, f=f): # … Read more
First thing you should consider when using recursion is the exit state; which means when the function will be exit public static int findMin(int[] numbers, int startIndex, int endIndex) { if(startIndex == endIndex){ return numbers[startIndex]; } int min = findMin(numbers, startIndex+1, endIndex); return numbers[startIndex] < min ? numbers[startIndex] : min ; } 0 solved Recursive … Read more
thanks to everyone who gave a answer 😀 much appreciated i got it working like this in the end int employmentInMonthsAmount = this.MaxEmploymentHistoryInMonths; var intcounter = 0; int years = 0; for (var i = 0; i < employmentInMonthsAmount; i++) { if (intcounter >= 12) { years++; intcounter = 0; } intcounter++; } var monthsleft … Read more
Here’s an example of how to do it. The idea is to iterate through your array, then look at every element after the one you’re currently looking at until you find one that is greater. The greater element (or -1 by default) is pushed into a second array for storage. var Arr1 = [4, 6, … Read more
The overall semantics of all three iteration statements is the same once they have been compiled into binary code. They just provide a different taste over the same thing. Performance depends on: the complexity of the body of the loop the complexity of calculating the end condition / next iteration step Since they all provide … Read more
You can use zip with a nested list comprehension: values = [10.5,20.2, 50.0] occ = [3,5,1] result = [x for x, y in zip(values, occ) for _ in range(y)] Alternatively, without using range: result = [elem for x, y in zip(values, occ) for elem in [x] * y] Output: [10.5, 10.5, 10.5, 20.2, 20.2, 20.2, … Read more
change scanf(“%s”, &input); to scanf(” %c”, &input); and get rid of getchar()s at the end. It will work as you wanted. %c specifier reads a char character, but it does not omit whitespace characters. Putting space onto scanf’s format will deal with whitespaces and read first non-whitespace char. To be more clear, see what man … Read more
for (int i = 1; i <= input; i++) { String line = “”; for (int k = 2; k >= 0; k–) { line += ((i >> k) & 1) == 1 ? “1” : “0”; } System.out.println(line); } That uses two for loops. solved Printing Binary values using nested loops [closed]
A do–while statement loops as long as the while expression is true. Your while expression is choice != ‘c’ || choice != ‘n’ In common English, that expression means choice is not ‘c’ OR choice is not ‘n’ That statement, logically, is always true. choice is always not one of those things. In both English … Read more
I don’t think you can avoid two loops, but you can make it sequential, instead of nested. long[] delta = new long[ar]; for(i=0;i<k;i++) { l=sc.nextLong(); r=sc.nextLong(); v=sc.nextLong(); if (l-1 < r) { m=(int)(l-1); delta[m] = v; m=(int)r; if (m < delta.length) { delta[m] = -v; } } } long cumulative = 0; for (int a … Read more
One particular use is to check even / odd: for($i=0; $i<10; $i++) { if($i%2==0) echo “even” else echo “odd” } This idea could be used to color rows (tr) of a table differently for better presentation. Another use is to close TR dynamically in a complex code (lets say, change tr after every 4 tds). … Read more
Your code is pretty fast (O(n)), but this problem can be solved in constant time (O(1)), since it is just based on the condition of the intersection of two lines being an integer. static String kangaroo(int k1, int v1, int k2, int v2) { float x = (k2 – k1)/(v1 – v2); if(x == (int) … Read more
You can use a for loop. range(5) loops 5 times whatever is inside it. The _ is to indicate that you don’t want to use the value returned by the iterator (In this case range(5)) for _ in range(5): if (value0 == 100): send_rckey(rc_exit) else: send_rckey(rc_program_up) value0 = checkPicture(15) For better understanding about for loop … Read more