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Are you supposed to use itoa() for this assignment? Because then you could use that to convert to a base 3 string, drop the last character, and then restore back to base 10.

The below code takes in 2 integers, and divides the first by the second. It supports negative numbers. int divide (int a, int b) { if (b == 0) //throw division by zero error //isPos is used to check whether the answer is positive or negative int isPos = 1; //if the signs are different, the answer will be negative if ((a < 0 && b > 0) || (a > 0 && b < 0)) int isPos = 0; a = Math.abs(a); b = Math.abs(b); int ans = 0; while (a >= b) { a = a-b; ans++; } if (isPos) return 0-ans; return ans; }

If you think about it you can divide by just using subtraction and a for() loop. Since you have this function itoa() available I'm assuming you are only using integers, which means you do not have to worry about remainders/decimals. Not gonna give you any code because you clearly just rewrote the assignment requirements, so give it a shot and if you have any questions post code and outline where you are having issues.

The function divides a 32-bit unsigned number by 3. If you multiply by 2^33 and then divide by 2^33 (by right shifting), then you get the original number. But if you multiply by (2^33)/3 and then divide by 2^33, you effectively divide by three. The last digit is B instead of A to cause the result to be rounded up. There is no need to actually write this in your code because the compiler will usually do this for you. Try it and see. (Furthermore, for a signed input, the compiler can safely generate a signed right shift, but the C language does not define such an operation.)

in Nimbers, in formula, keyboard don't allowse me to type this sign - / Why?

If this is what you need (numbers increase columnwise but rows need to be filled first): for 16: 0 6 11 1 7 12 2 8 13 3 9 14 4 10 15 5 for 17: 0 6 12 1 7 13 2 8 14 3 9 15 4 10 16 5 11 You need to define which column is getting the remainders: int remainder = total % 3; if remainder is 1, only first column is 6 elements. if remainder is 2, first & second columns are 6 elements. You need to calculate startIndex and endIndex according to this. So; int pageSize = total / 3; int remainder = total % 3; int startIndex = column * pageSize + min(column, remainder); int endIndex = startIndex + pageSize + (remainder > column ? 1 : 0); should work. I just tested it, it works for different rowsizes than 3. Here is how I got the formulas, drawing tables is good practice for sorting out such algortihms: r:Remainder, c:column, ps:pagesize (as calculated above) StartingIndex: . |r:0 |r:1 |r:2 ---------------------- c:0|0 |0 |0 ---------------------- c:1|ps |ps+1 |ps+1 ----------------------…

• The most straightforward solution is to use modulo 3 to check for divisibility, but that's not going to be recursive. • An alternate solution is to recursively keep dividing by 3 until you get down to 1, but that will stack overflow for large values. • A third solution, which lends itself to recursion, is to leverage the property that if the sum of the digits of a number is divisible by 3, then the number is divisible by 3. Here's an implementation of the third option that avoids modulo arithmetic and copes with very large numbers: def divThree(num): if num < 10: return (num in [3, 6, 9]) else: return divThree(sum([int(digit) for digit in str(num)])) You can add 0 to the list in the first return if you want to consider it divisible by 3 as well. If you want to accommodate both positive and negative values, prepend: if num < 0: return divThree(-num) as the first check to be performed.

You should have a variable to sum numbers which satisfies condition numberArray[i] % 3 == 0: function sumNumbersBy3(numberArray) { let sum = 0 for(let i = 0; i < numberArray.length; i++) { if (numberArray[i] % 3 == 0) sum += numberArray[i]; } return sum; } In addition, you can use reduce method: const result = arr.reduce((a, c, i) => { (c % 3 == 0) ? a += c : 0; return a; }, 0); An example: function sumNumbersBy3(numberArray) { let sum = 0 for(let i = 0; i < numberArray.length; i++) { if (numberArray[i] % 3 == 0) sum += numberArray[i]; } return sum; } console.log("sumNumbersBy3: ", sumNumbersBy3([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])); let arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]; const result = arr.reduce((a, c, i) => { (c % 3 == 0) ? a += c : 0; return a; }, 0); console .log(`Using reduce method: ${result}`); Run code snippetHide resultsExpand snippet

I think you need something like this int divide(int a, int b){ if(a - b <= 0){ return 1; } else { return divide(a - b, b) + 1; } }

This version takes precision as a parameter as well: with q as (select 100 as val, 3 as parts, 2 as prec from dual) select rownum as no ,case when rownum = parts then val - round(val / parts, prec) * (parts - 1) else round(val / parts, prec) end v from q connect by level <= parts no v === ===== 1 33.33 2 33.33 3 33.34 For example, if you want to split the value among the number of days in the current month, you can do this: with q as (select 100 as val ,extract(day from last_day(sysdate)) as parts ,2 as prec from dual) select rownum as no ,case when rownum = parts then val - round(val / parts, prec) * (parts - 1) else round(val / parts, prec) end v from q connect by level <= parts; 1 3.33 2 3.33 3 3.33 4 3.33 ... 27 3.33 28 3.33 29 3.33 30 3.43 To apportion the value amongst each month, weighted by the number of days in each month, you could do this instead (change the level <= 3 to change the number of months it is calculated for): with q as ( select add_months(date '2013-07-01…