The given code reads an integer from the standard input, calculates the triangular numbers up to that integer, and outputs them. Triangular numbers are the sums of natural numbers up to a given number.

Here's a breakdown of the code:

  1. Read the integer input from the standard input using intval(fgets(STDIN)) and store it in the variable $n.

  2. Calculate the block size as the integer square root of $n using intval(sqrt($n)) and store it in the variable $block.

  3. Initialize an empty string variable $output to store the output.

  4. Initialize a counter variable $sn to keep track of the number of values added to the output.

  5. Iterate over each block of numbers. For each block:

    • Calculate the start and end values for the block using the block index and block size.
    • Iterate over the numbers in the block.
    • Calculate the triangular number for each number using the formula (($i + 1) * $i) >> 1 (bitwise right shift by 1 is equivalent to integer division by 2).
    • Append the calculated value to the $output string.
    • Increment the counter $sn.
    • If the counter reaches 500,000, output the current $output string and reset the counter and $output to empty.
  6. Calculate the start and end values for the last block outside the loop using the block size and the remaining numbers.

    • Iterate over the numbers in the last block.
    • Calculate the triangular number for each number and append it to the $output string.
    • Increment the counter $sn.
    • If the counter reaches 500,000, output the current $output string and reset the counter and $output to empty.
  7. Finally, output the remaining values in the $output string.

Note: The code is dividing the numbers by blocks and outputting them in chunks of 500,000 values. This is done to avoid memory overflow when dealing with large inputs

php$n = intvalfgetsSTDIN;$block = intvalsqrt$n;$output = ;$sn=0;for $b = 0; $b $block; $b++ 	$start = $b $block + 1;	$end = $b + 1 $block + 1;	for $i = $start; $i $end; $i++ 		$value = $i + 1 $i

原文地址: http://www.cveoy.top/t/topic/iSJv 著作权归作者所有。请勿转载和采集!

免费AI点我,无需注册和登录