project euler 53

Problem 53

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Combinatoric selections

There are exactly ten ways of selecting three from five, 12345:

123, 124, 125, 134, 135, 145, 234, 235, 245, and 345

In combinatorics, we use the notation, 5C3 = 10.

In general,

nCr=n!r!(n−r)!, where r ≤ n, n! = n×(n−1)×…×3×2×1, and 0! = 1.

It is not until n = 23, that a value exceeds one-million: 23C10 = 1144066.

How many, not necessarily distinct, values of  nCr, for 1 ≤ n ≤ 100, are greater than one-million?

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组合数选择

从五个数12345中选择三个恰好有十种方式，分别是：

123、124、125、134、135、145、234、235、245和345

在组合数学中，我们记作：5C3 = 10。

一般来说，

nCr=n!r!(n−r)!，其中r ≤ n，n! = n×(n−1)×…×3×2×1，且0! = 1。

直到n = 23时，才出现了超出一百万的组合数：23C10 = 1144066。

若数值相等形式不同也视为不同，对于1 ≤ n ≤ 100，有多少个组合数nCr超过一百万？

package cuppics;

import java.util.Arrays;

import junit.framework.TestCase;

public class Prj53 extends TestCase {

	public void testCombinatricSelections() {

		calculateCombinaticSelections(100);
	}

	static final int UP_LIMIT = 1000000;
	static final int LIMIT_MASK = Integer.MAX_VALUE;

	public static int[] calculateCombinaticSelections(int num) {

		assert (num >= 1);

		if (num == 1) {
			return new int[] { 1, 1 };
		} else if (num == 2) {
			return new int[] { 1, 2, 1 };
		}

		int[] tmp = calculateCombinaticSelections(2);

		int numOfBigThanLimit = 0;

		for (int i = 3; i <= num; i++) {

			int[] val = new int[i + 1];

			for (int j = 0; j <= i / 2; j++) {
				if (j == 0) {
					val[j] = 1;
					continue;
				}

				if (tmp[j - 1] == LIMIT_MASK || tmp[j] == LIMIT_MASK) {
					val[j] = LIMIT_MASK;

					numOfBigThanLimit = updateCount(numOfBigThanLimit, i, j);

					break;
				} else {

					val[j] = tmp[j - 1] + tmp[j];

					if (j == i / 2 && i % 2 == 0) {
						val[j] = tmp[j - 1] * 2;
					}

					if (val[j] >= UP_LIMIT) {
						val[j] = LIMIT_MASK;
						numOfBigThanLimit = updateCount(numOfBigThanLimit, i, j);
						break;

					}
				}

			}

			tmp = Arrays.copyOf(val, val.length);

		}
		return tmp;
	}

	static int updateCount(int numOfBigThanLimit, int i, int j) {
		int numOf = (i + 1) / 2 - j;
		numOf *= 2;
		numOf = (i % 2 != 0) ? numOf : (numOf + 1);
		numOfBigThanLimit += numOf;
		return numOfBigThanLimit;
	}

}