ABC - 108- A&B&C

AtCoder ABC 108 A&B&C

AtCoder - 108

A問題

• ループしてカウントしたパターン

	private void solveA() {
		Scanner scanner = null;
		int numK = 0;
		int eqaC = 0;
		int oddC = 0;

		try {
			scanner = new Scanner(System.in);
			numK = scanner.nextInt();
			for (int i = 1; i <= numK; i++) {
				if (i % 2 == 0) {
					eqaC++;
				} else {
					oddC++;
				}
			}

			System.out.println(eqaC * oddC);

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

A問題:2

• 数値K
  • 奇数の個数：$(K+1)/2$
  • 偶数の個数：$K/2$

	private void solveA2() {
		Scanner scanner = null;
		int numK = 0;

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

			System.out.println((numK / 2) * ((numK + 1) / 2));

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

B問題

• 頂点が反時計回りの長方形
  • 紙に実際に書いてみると頂点の算出は簡単（頭の中だけだとﾁｮｲ無理）

	private void solveB() {
		Scanner scanner = null;
		int x1 = 0;
		int y1 = 0;
		int x2 = 0;
		int y2 = 0;

		try {
			scanner = new Scanner(System.in);
			x1 = scanner.nextInt();
			y1 = scanner.nextInt();
			x2 = scanner.nextInt();
			y2 = scanner.nextInt();

			int x3 = x2 - (y2 - y1);
			int y3 = y2 + (x2 - x1);
			int x4 = x3 - (y3 - y2);
			int y4 = y3 + (x3 - x2);

			System.out.println(x3 + " " + y3 + " " + x4 + " " + y4);

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

C問題

上手くまとめられないので思考過程を記述するのを放棄する

参考：これらのサイトのお世話になりました
ARC102[Triangular Relationship]
ABC108: C - Triangular Relationship
AtCoder ARC 102 C - Triangular Relationship (300 点)

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

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

			long cnt = 0;

			if (numK % 2 != 0) {
				long numCount = 0;
				for (int i = 1; i <= numN; i++) {
					if (i % numK == 0) {
						numCount++;
					}
				}
				cnt = (long) Math.pow(numCount, 3);
			} else {
				long numCount = 0;
				long numCount2 = 0;
				for (int i = 1; i <= numN; i++) {
					if (i % numK - numK / 2 == 0) {
						numCount++;
						//						System.out.println(i + " : 0");
					} else if (i % numK == 0) {
						numCount2++;
						//						System.out.println(i + " : 1");
					}
				}
				cnt = (long) Math.pow(numCount, 3) + (long) Math.pow(numCount2, 3);
			}

			System.out.println(cnt);

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