\n The problem of finding the sum of consecutive natural numbers is a common topic in coding tests. \n – Integer \n – Output \n To solve this problem, it is useful to first rewrite the condition \n The above equation can be transformed to \n 1. To find possible \n Below is the code that implements the above logic in Kotlin.\n <\/p>\n \n – The
\n Given the input integer N<\/code>, the problem is to find M<\/code> such that the sum of consecutive natural numbers from 1<\/code> to M<\/code> equals N<\/code>. Here, M<\/code> must be a natural number greater than or equal to 1, and the sum of consecutive natural numbers can be calculated using the following formula:\n <\/p>\nS = 1 + 2 + 3 + ... + M = M * (M + 1) \/ 2<\/code><\/pre>\nInput<\/h2>\n
N<\/code> (1 \u2264 N<\/code> \u2264 1,000,000) – the sum of natural numbers to be found\n <\/p>\nOutput<\/h2>\n
M<\/code> when the sum of consecutive natural numbers from 1<\/code> to M<\/code> equals N<\/code>.
\n – If M<\/code> does not exist, output -1<\/code>.\n <\/p>\nApproach to Problem<\/h2>\n
S = N<\/code> into an equation.
\n When setting S = 1 + 2 + ... + M<\/code> equal to N<\/code>, rearranging gives us:\n <\/p>\nN = M * (M + 1) \/ 2<\/code><\/pre>\n2N = M * (M + 1)<\/code>.
\n We can then solve for M<\/code> using this equation and check if the value is a natural number.\n <\/p>\nSolution Process<\/h2>\n
M<\/code> for N<\/code>, start with M<\/code> equal to 1 and increase it while calculating S<\/code>.
\n 2. If S<\/code> equals N<\/code>, print M<\/code> and terminate.
\n 3. If S<\/code> exceeds N<\/code>, there is no need to search further, print -1<\/code> and terminate.\n <\/p>\nCode Implementation<\/h2>\n
fun findContinuousNaturalSum(N: Int): Int {\n var sum = 0\n for (M in 1..N) {\n sum += M\n if (sum == N) {\n return M\n } else if (sum > N) {\n return -1\n }\n }\n return -1\n }\n\n fun main() {\n val N = 15\n val result = findContinuousNaturalSum(N)\n println(\"The value of M such that the sum of consecutive natural numbers equals $N is: $result\")\n }<\/code><\/pre>\nCode Explanation<\/h2>\n
findContinuousNaturalSum<\/code> function finds M<\/code> such that the sum of consecutive natural numbers equals
\n N<\/code>.
\n – The sum<\/code> variable stores the sum from 1<\/code> to M<\/code>.
\n – Using a for<\/code> loop, M<\/code> is incremented from 1<\/code> to N<\/code>, while calculating sum<\/code>.
\n – If sum<\/code> equals N<\/code>, M<\/code> is returned, and if sum<\/code> exceeds N<\/code>, -1<\/code> is returned to terminate.\n <\/p>\nExamples<\/h2>\n