ABC - 095- A&B&C

AtCoder ABC 095 A&B&C

AtCoder - 095

A問題

• 文字列を1つずつ○×判定

	private void solveA() {
		Scanner scanner = null;
		String topping = "";

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

			int res = 700;

			for (int i = 0; i < topping.length(); i++) {
				switch (topping.charAt(i)) {
				case 'o':
					res += 100;
					break;
				case 'x':
					break;

				}
			}

			System.out.println(res);

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

B問題

• N 種類のドーナツそれぞれを少なくとも 1 個は作る。という制約よりN種類のドーナツを1つずつ作ることは可能
  • Xから、N種類のドーナツ全て最低1回作成したとして $m_i$ を引く
  • $m_i$ が一番低いドーナツをたくさん作りたいので、ソートして、Xの残りを全て $m_i$ にあてる

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

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

			int[] wk = new int[numN];

			int res = 0;
			for (int i = 0; i < numN; i++) {
				wk[i] = scanner.nextInt();
				res++;
				numX -= wk[i];
			}

			Arrays.sort(wk);

			boolean isExit = false;
			while (!isExit) {
				if (numX >= wk[0]) {
					numX -= wk[0];
					res++;
				} else {
					isExit = true;
				}
			}

			System.out.println(res);

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

C問題

• ピザの枚数は $2×C= A ＋ B$ でも同じものを作れる
• $2×C$ と $ A ＋ B$ のどちらが安いかを判定
  • $2×C$ と $ A $ のどちらが安いかを判定
  • $2×C$ と $ B $ のどちらが安いかを判定

	private void solveC() {
		Scanner scanner = null;
		int numA = 0;
		int numB = 0;
		int numC = 0;
		int numX = 0;
		int numY = 0;

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

			int sabun1 = Math.min(numX, numY);

			int res = 0;
			if (numA + numB >= numC * 2) {
				res += sabun1 * (numC * 2);
			} else if (numA + numB < numC * 2) {
				res += sabun1 * (numA + numB);
			}

			if (numX > numY) {
				if (numA <= numC * 2) {
					res += (numX - sabun1) * numA;
				} else {
					res += (numX - sabun1) * (numC * 2);
				}
			} else if (numX < numY) {
				if (numB <= numC * 2) {
					res += (numY - sabun1) * numB;
				} else {
					res += (numY - sabun1) * (numC * 2);
				}
			}

			System.out.println(res);

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