ABC - 070- A&B&C

AtCoder ABC 070 A&B&C

AtCoder - 070

A問題

• 数値の比較ではなく文字列で比較。StringBuilder#reverseで逆順に変換
  • 便利なんだけど速度は遅い

	private void solveA() {
		String numN = next();

		out.println(numN.equals(new StringBuilder(numN).reverse().toString()) ? "Yes" : "No");
	}

B問題

• A,B,C,Dの組み合わせは以下になる
  • （A↔C）（B↔D）の位置は変換可
  • ● の位置が重なる秒数（と思って）

---	---	---	---	---
A	●	●	●	B
C	●	●	●	D
---	---	---	---	---
A		●	B
	C	●		D
---	---	---	---	---
	A	●	B
C		●		D
---	---	---	---	---
A	B
			C	D

	private void solveB() {
		int numA = nextInt();
		int numB = nextInt();
		int numC = nextInt();
		int numD = nextInt();

		int wkRes = Math.min(numB, numD) - Math.max(numA, numC);
		//wkReが0より小さい場合、二人がスイッチを押していた時間の被りはない
		out.println(wkRes >= 0 ? wkRes : 0);
	}

C問題

• この時計達の最小公倍数を求めれば、もう一度すべての針が上を向く
  • 最小公倍数=$(val1 * val2)/(val1とval2の最大公約数)$

	private void solveC() {
		int numN = nextInt();

		BigDecimal[] times = new BigDecimal[numN];
		for (int i = 0; i < numN; i++) {
			times[i] = new BigDecimal(next());
		}


		/*
		 * 1つしか要素がない場合は要素の秒数がかかる秒数
		 */
		if (times.length == 1) {
			out.println(times[0]);
			return;
		}

		/*
		 * 最小公倍数を求める
		 */
		BigDecimal res = BigDecimal.ZERO;
		for (int i = 1; i < times.length; i++) {
			if (i == 1) {
				res = getMinKoubaisu(times[i - 1], times[i]);
			} else {
				res = getMinKoubaisu(res, times[i]);
			}
		}

		out.println(res.longValue());
	}

	private BigDecimal getMinKoubaisu(BigDecimal val1, BigDecimal val2) {
		BigDecimal wk = getMaxKouyaku(val1, val2);
		return (val1.multiply(val2).divide(wk));
	}

	private BigDecimal getMaxKouyaku(BigDecimal val1, BigDecimal val2) {

		BigDecimal wkVal1 = val1.compareTo(val2) <= 0 ? val1 : val2;
		BigDecimal wkVal2 = val1.compareTo(val2) > 0 ? val1 : val2;
		BigDecimal amari = wkVal2.remainder(wkVal1);
		while (!amari.equals(BigDecimal.ZERO)) {
			wkVal2 = wkVal1;
			wkVal1 = amari;
			amari = wkVal2.remainder(wkVal1);
		}

		return wkVal1;
	}