ABC - 086- A&B&C

AtCoder ABC 086 A&B&C

AtCoder - 086

A問題

    private void solveA() {
        Scanner scanner = null;
        int numA = 0;
        int numB = 0;

        try {
            scanner = new Scanner(System.in);
            numA = scanner.nextInt();
            numB = scanner.nextInt();

            if (numA * numB % 2 == 0) {
                System.out.println("Even");
            } else {
                System.out.println("Odd");
            }

        } finally {
            if (scanner != null) {
                scanner.close();
            }
        }
    }

B問題

解法1

• 文字列として取得
• 繋げてsqrt
• intにキャストして整数にする
• 二乗して元の数値になればOK

解法2

• 数値は最大でも100100なので、$0^2$ - $1000^2$　まで総当たりで調べる手もあるかな

    private void solveB() {
        Scanner scanner = null;
        String numA = "";
        String numB = "";

        try {
            scanner = new Scanner(System.in);
            numA = scanner.next();
            numB = scanner.next();

            double wkNum = Double.parseDouble(numA + numB);
            int wk = (int) Math.sqrt(wkNum);
            if (Math.pow(wk, 2) == wkNum) {
                System.out.println("Yes");
            } else {
                System.out.println("No");
            }

        } finally {
            if (scanner != null) {
                scanner.close();
            }
        }
    }

C問題

• 時刻0秒時に$(0,0)$からスタート
• $T_i$秒後に$(X_i,Y_i)$点にいる場合
  • $T_i$秒が偶数なら、$(X_i+Y_i)$は偶数である
    • なぜなら、1秒間で動けるのは以下の4パターンのみ
      • $(X_i+1,Y_i)$ - $(X_i,Y_i+1)$
      • $(X_i-1,Y_i)$ - $(X_i,Y_i-1)$
    • そのため、$T_i$秒が偶数なら、$(X_i+Y_i)$の偶奇は一致する
  • $経過時間T \geqq (Xの移動量+Yの移動量)$は常に成り立つ

private void solveC() {
        Scanner scanner = null;
        int numN = 0;

        try {
            scanner = new Scanner(System.in);
            numN = scanner.nextInt();
            int[] t = new int[numN];
            int[] x = new int[numN];
            int[] y = new int[numN];

            for (int i = 0; i < t.length; i++) {
                t[i] = scanner.nextInt();
                x[i] = scanner.nextInt();
                y[i] = scanner.nextInt();
            }

            int cT = 0;
            int cX = 0;
            int cY = 0;
            for (int i = 0; i < numN; i++) {

                boolean wkB = (t[i] % 2 == 0);
                boolean wkB2 = ((x[i] + y[i]) % 2 == 0);

                if (wkB != wkB2) {
                    System.out.println("No");
                    return;
                }

                if (t[i] - cT < Math.abs(Math.abs(cX + cY) - Math.abs(x[i] + y[i]))) {
                    System.out.println("No");
                    return;
                }
                cT = t[i];
                cX = x[i];
                cY = y[i];
            }

            System.out.println("Yes");

        } finally {
            if (scanner != null) {
                scanner.close();
            }
        }
    }