AtCoder ABC 083 A&B&C

AtCoder - 083

A問題

• 比較

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

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

			if (numA + numB == numC + numD) {
				System.out.println("Balanced");
			} else if (numA + numB > numC + numD) {
				System.out.println("Left");
			} else if (numA + numB < numC + numD) {
				System.out.println("Right");
			}

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

B問題

• ただ単純に、1－Nまで各桁の和を調べ、 $A \geqq 各桁の和 \geqq B$となる数値を出力
  • 12 -> 1+2=3
• 各桁の和は $mod10$ , $divide 10$を繰り返せばよい(modの値を+していく)
  • $12 \quad \rightarrow \quad mod 10 = 2 \quad \rightarrow \quad 12/10 = 1 \quad \rightarrow \quad mod10=1$

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

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

			int res = 0;
			for (int i = 1; i <= numN; i++) {

				int wk = 0;
				int tmp = i;
				while (tmp != 0) {
					wk += tmp % 10;
					tmp /= 10;
				}
				if (numA <= wk && wk <= numB) {
					res += i;
				}

			}

			System.out.println(res);

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

C問題

　- $A_{i+1}はA_iの倍数$ということは、$A_1はA_0の倍数$
- $A_0はXの倍数 \rightarrow A_0 = X$
- $A_{i+1}はA_iの倍数 \rightarrow A_{i+1}はXの倍数$
- 数列の長さの最大値なので、倍数が最小であれば良い →　倍数は2が最小（1だと条件を満たさない）
- $X*2n \leqq Y$ 以下になる $nの最大値$ を求める

	private void solveC() {
		Scanner scanner = null;
		long numX = 0;
		long numY = 0;

		try {
			scanner = new Scanner(System.in);
			numX = scanner.nextLong();
			numY = scanner.nextLong();

			long x = numX;
			int res = 0;
			while (x <= numY) {
				x *= 2;
				res++;
			}

			System.out.println(res);

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