ABC - 041- A&B&C

AtCoder ABC 041 A&B&C

AtCoder - 041

A問題

• 文字列切り出し

	private void solveA() {
		String wk = next();
		int numI = nextInt();

		out.println(wk.substring(numI - 1, numI));
	}

B問題

• 剰余の性質
  • $(A × B) ; mod ; M = ((A ; mod ; M) × (B ; mod ; M)) ; mod ; M$

	private void solveB() {
		long numA = nextLong();
		long numB = nextLong();
		long numC = nextLong();

		long CONST_MOD = (long) (Math.pow(10, 9) + 7);

		long res = numA * numB % CONST_MOD * numC % CONST_MOD;

		out.println(res);
	}

C問題

• 二次元配列をソートするだけなのでjavaだとすごく簡単
• 次はGOでこの実装を試す

	private void solveC() {
		int numN = nextInt();
		int[][] wk = IntStream.range(0, numN).collect(() -> new int[numN][2],
				(t, i) -> {
					t[i][0] = i + 1;
					t[i][1] = nextInt();
				}, (t, u) -> {
					Stream.concat(Arrays.stream(t), Arrays.stream(u));
				});

		Arrays.sort(wk, (x, y) -> {
			if (x[1] < y[1]) {
				return 1;
			} else if (x[1] > y[1]) {
				return -1;
			} else {
				if (x[0] < y[0]) {
					return -1;
				} else if (x[0] > y[0]) {
					return 1;
				} else {
					return 0;
				}
			}
		});

		for (int j = 0; j < wk.length; j++) {
			out.println(wk[j][0]);
		}
	}