给定数字是否可以写成两个或多个连续正整数的总和？

Question

I need to write a method which takes in an int and returns true if the number can be written as a sum of two or more consecutive positive integers and false otherwise.

boolean IsSumOfConsecutiveInts(int num)

I figured out that all odd numbers (except the number 1) can be written as the sum of 2 consecutive positive integers:

return (num > 1 && num % 2 == 1);

but this doesn't account for numbers that can be written as the sum of more than 2 consecutive positive integers (such as 6 == 1 + 2 + 3 ).

How can I determine whether a number can be written as a sum of two or more consecutive positive integers?

Answer 1

These numbers are called Polite Numbers .

And, conveniently, the only numbers that aren't polite are the powers of 2.

So, that gives us 2 options. We can either determine that a number is polite , OR we can determine that it is not a power of 2 .

I did both; the latter is easier (and more efficient).

1. This determines whether or not a number is polite :

    boolean IsSumOfConsecutiveInts(int num) { int sumOfFirstIIntegers = 3; for (int i = 2; sumOfFirstIIntegers <= num; i++) { if (i%2 == 0 ? (num%i == i/2) : (num%i == 0)) { return true; } sumOfFirstIIntegers += i + 1; } return false; } 

   This one is pretty hard to understand. It took me a while to come up with.

   Basically, i is the number of consecutive integers that we are checking;

   sumOfFirstIIntegers is equal to the sum of the first i integers, so that means that all the numbers that can be expressed as a sum of i consecutive integers are greater than or equal to sumOfFirstIIntegers .

   The last part that deserves discussing is the boolean statement i%2 == 0 ? (num%i == i/2) : (num%i == 0) i%2 == 0 ? (num%i == i/2) : (num%i == 0) . Let's look at some examples:

    i all sums of i consecutive positive integers 2 3, 5, 7, 9... 3 6, 9, 12, 15... 4 10, 14, 18, 22... 5 15, 20, 25, 30... 

   There are two cases, but in either case, we can express all possible numbers that are a sum of i consecutive integers pretty simply.

   1. When i is even, num must be equal to (i * n) + (i / 2) where n is a non-negative integer. This can of course be written as num % i == i / 2 .

   2. When i is odd, num must be equal to i * n , where n is a non-negative integer. Which gives us our second condition num % i == 0 .

   In addition to these conditions, num can not be less than the sum of the first i positive integers. Hence, our for loop's conditional: sumOfFirstIIntegers <= num .

2. This determines whether a number is not a power of 2 :

    boolean IsSumOfConsecutiveInts(int num) { return (num & (num - 1)) != 0; } 

   This answer does a good job of explaining why this works.

Note that both of the above solutions have the same result, they are just different ways of thinking about the problem.