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関数calcXと関数calcYは，引数inDataを用いて計算を行い，その結果を戻り値とする。関数calcXをcalcX(1)として呼び出すと，関数calcXの変数numの値が，1→3→7→13と変化し，戻り値は13となった。関数calcYをcalcY(1)として呼び出すと，関数calcYの変数numの値が，1→5→13→25と変化し，戻り値は25となった。プログラム中のa，bに入れる字句の適切な組合せはどれか。 〔プログラム 1〕 ○整数型: calcX(整数型: inData) 整数型: num, i num ← inData for (i を 1 から 3 まで 1 ずつ増やす) num ← a endfor return num 〔プログラム 2〕 ○整数型: calcY(整数型: inData) 整数型: num, i num ← inData for ( b ) num ← a endfor return num
Here a is 2×num+i. Substituting it into calcX, i=1 gives 2×1+1=3 which matches, but i=2 gives 2×3+2=8, not the required 7. The expression for a is wrong and cannot reproduce the change sequence 1→3→7→13.
Here a is 2×num+i, which like ア cannot reproduce calcX (i=2 yields 8 instead of 7). Although the range in b is correct, the combination is unsuitable because a is wrong.
a = num+2×i is correct, but with b as “i from 1 to 7, increasing by 3 each time”, i = 1, 4, 7 — three times — and the first step i=1 gives 1+2×1=3, not the required 5, so it cannot reproduce calcY's 1→5→13→25. b is wrong.
正解 First determine a using calcX. calcX(1) starts num at 1, and the for loop moves the loop variable i as 1→2→3, three times in all. Substituting a = num+2×i gives 1+2×1=3 at i=1, 3+2×2=7 at i=2, and 7+2×3=13 at i=3, matching the change sequence 1→3→7→13 and the return value 13. Hence a = num+2×i. Next determine b using calcY. calcY uses the same a = num+2×i and changes num as 1→5→13→25 over three steps, so its loop must also run three times. If b is “i from 2 to 6, increasing by 2 each time”, then i = 2, 4, 6 — three times — giving 1+2×2=5 at i=2, 5+2×4=13 at i=4, and 13+2×6=25 at i=6, which matches. Therefore the combination エ, with a = num+2×i and b = “i from 2 to 6, increasing by 2 each time”, is correct.