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最新推荐文章于 2024-12-08 11:08:43 发布

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本文探讨了非弹跳数的概念，即既不递增也不递减的正整数，并通过算法计算了特定范围内非弹跳数的数量。给出了针对不同数量级（如10^100）的非弹跳数的具体计算方法。

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Non-bouncy numbers

Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, 134468.

Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, 66420.

We shall call a positive integer that is neither increasing nor decreasing a “bouncy” number; for example, 155349.

As n increases, the proportion of bouncy numbers below n increases such that there are only 12951 numbers below one-million that are not bouncy and only 277032 non-bouncy numbers below 1010.

How many numbers below a googol (10100) are not bouncy?

非弹跳数

从左往右，如果每一位数字都大于等于其左边的数字，这样的数被称为上升数，比如134468。

同样地，如果每一位数字都大于等于其右边的数字，这样的数被称为下降数，比如66420。

如果一个正整数既不是上升数也不是下降数，我们就称之为“弹跳”数，比如155349。

随着n的增长，小于n的弹跳数的比例也随之增长；在小于一百万的数中，只有12951个非弹跳数，而小于1010的数中只有277032个非弹跳数。

在小于一古戈尔（10100）的数中有多少个非弹跳数？

package projecteuler;

import java.math.BigInteger;
import junit.framework.TestCase;

public class Prj113 extends TestCase {


	/**[a1,a2,a3,....an],  1<=a1<=a2<=a3<=...an<=9
	 *   map to ---> [d + i -1]
	 * one to one map ===>rC(n+r-1)
	 */
	public void testNonBouncyNumbers() {

		assertEquals(calNumOfNonbouncy(10, 6).toString(), "12951");
		assertEquals(calNumOfNonbouncy(10, 10).toString(), "277032");
		System.out.println(calNumOfNonbouncy(10, 100));
	}

	BigInteger calNumOfNonbouncy(int _n, int _r) {

		BigInteger ret = BigInteger.ZERO;

		int n = _n;
		int r = _r;
		BigInteger asc = calNr(n + r - 1, r);
		// zero out
		ret = ret.add(asc).subtract(BigInteger.ONE);

		// [9,8,7,6,5,4,3,2,1]

		n = n - 1;
		//倒序补上0
		for (int i = 1; i <= r; i++) {
			BigInteger desc = calNr(n + i - 1, i).multiply(
					new BigInteger(Integer.toString(r + 1 - i)));
			ret = ret.add(desc);
		}

		ret = ret.subtract(new BigInteger(Integer.toString(_r * (_n - 1))));

		return ret;
	}

	BigInteger calNr(int _n, int _r) {

		BigInteger ret = BigInteger.ONE;
		for (int i = 1; i <= _r; i++) {
			ret = ret.multiply(new BigInteger(Integer.toString(_n + 1 - i)));
		}
		for (int i = 1; i <= _r; i++) {
			ret = ret.divide(new BigInteger(Integer.toString(i)));
		}

		return ret;
	}

}