In an if else-if statement you put multiple conditions to evaluate the result. The following is how the statements will work in your case: if(row==1||row==n||row==2*n-1) cout<<“*”; //if true then put * or if false then move down else if (col==1&&row<n||col==n&&row>n) cout<<“*”; // if the first is false and the 2nd one is true then put … Read more
In your outer while loop you need this: overtime = 0; if (hoursWorked>8) overtime = (hoursWorked-8)*(rateOfPay*1.5) – (hoursWorked-8)*rateOfPay; // Calculate OVERTIME HOURS sets the overtime to zero for each iteration, and you’ll have to subtract the normal rate for overtime hours (otherwise you add them twice) and then you’ll have to add the overtime to … Read more
Suppose there are n people attending this quiz, the # of the winner is x. So the sum of people entering before the winner is the sum of 1+…+(x-1), the sum of people entering after the winner is the sum of (x+1)+…+n. Here we have 1+…+(x-1) = (x+1)+…+n, which derives to (x*(x-1))/2 = ((x+n+1)*(n-x))/2. Simplifying … Read more
Its a simple math sum quiz that is asking the user for sum of same numbers from 1 to 10 The expected output is : What is 1 + 1? In every single iteration of the outer loop , after questioning like this , the program waits for the user input. When the user inputs … Read more
You have put the k=k+1 at a wrong place. It is the correct code: stab1=zeros(1,4991); k=0; for ii=-0.6:0.01:-.3 m=0; for jj=0:0.01:1.6 m=m+1; if …. (some condition) stab1(k*161+m)=1; end end k=k+1; end Now it has 4991 elements. solved How many elements the output matrix should have in this code? [closed]
just replace the condition of the there loops with 26. Now the exception index out of bound will not occur public class SuggestingAppNames { public static void main(String[] args) { System.out.println(“the possible outcomes are”); String a = “A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z”; String d[] = a.split(“,”); String b = “A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z”; String e[] = b.split(“,”); String c = “A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z”; String … Read more
When I include brackets in a nested for loop, it doesn’t function correctly but when I take them out it does; why is this? [closed] solved When I include brackets in a nested for loop, it doesn’t function correctly but when I take them out it does; why is this? [closed]
You need for-loop with if/else to run different code for different lists. For list with “PRODUCT NAME PACK” you have to keep list with names in variable – so you may use it in next loops when you get list with numbers For other list you can keep first number as index and rest use … Read more
Based on provided results, you need updated first loop and second. Fist loop should start from 9 as you already printed mid line in previous part. And second loop should iterate to counter of fist loop to achieve what you except: public static void main(String[] args) { for (int i = 10; i >= 1; … Read more
Maybe use zip: for a,b,c in zip(scores,boxes,class_list): if a >= 0.98: data[k+1] = {“score”:a,”box”:b, “class”: c } ; k=k+1; print(“Final_result”,data) Output: Final_result {2: {‘score’: 0.98, ‘box’: [0.1, 0.2, 0.3, 0.4], ‘class’: 1}} Edit: for a,b,c in zip(scores,boxes,class_list): if a >= 0.98: data[k+1] = {“score”:a,”box”:b, “class”: int(c) } ; k=k+1; print(“Final_result”,data) 9 solved Appeding different list … Read more
I am not sure what exactly you are stuck on based on your question so I made a minimal C program with comments I declared an int array who’s first and second dimension are at least 10 because you iterate both i and j from 0 to 9 (inclusive). This is to avoid out of … Read more
This would be the equivalent in python. In range(0, height, cellSize), 0 and height are the bounds of the range, and cellSize is how many out counter increments. for y in range(0, height, cellSize): for x in range(0, width, cellSize): rect(x, 0, cellSize, cellSize) 5 solved Python single line for loops [duplicate]