This question's explanation and your answers so far.
手続 printArray は，配列 integerArray の要素を並べ替えて出力する。手続 printArray を呼び出したときの出力はどれか。ここで，配列の要素番号は 1 から始まる。 〔プログラム〕 ○printArray() 整数型: n, m 整数型の配列: integerArray ← {2, 4, 1, 3} for (n を 1 から (integerArray の要素数 − 1) まで 1 ずつ増やす) for (m を 1 から (integerArray の要素数 − n) まで 1 ずつ増やす) if (integerArray[m] > integerArray[m + 1]) integerArray[m] と integerArray[m + 1] の値を入れ替える endif endfor endfor integerArray の全ての要素 を先頭から順にコンマ区切りで出力する
正解 This program compares two adjacent elements and swaps them when the left one is greater than the right one — a bubble sort (ascending order). The initial array is {2, 4, 1, 3}. The outer variable n runs from 1 to 3, and the inner variable m runs from 1 to (4 − n). In the n=1 pass (m=1 to 3), {2,4,1,3}→{2,1,4,3}→{2,1,3,4}; in the n=2 pass (m=1 to 2), {2,1,3,4}→{1,2,3,4}; in the n=3 pass (m=1) no swap occurs, so it stays {1,2,3,4}. Therefore the output is 1, 2, 3, 4, and the correct answer is ア.
1, 3, 2, 4 has 2 and 3 in reversed order, so it is not sorted in ascending order. When each pass is carried out to the end, 2 comes before 3, so this arrangement does not occur.
3, 1, 4, 2 is not sorted; it is a wrong result that processes the initial array only partway (or mistakes the comparison condition). The program performs all passes and sorts completely into ascending order.
4, 3, 2, 1 is descending order. The condition of this program is to swap when the left value is greater than the right value, which is an ascending sort that moves larger values toward the back, so it does not produce descending order.