TopCoder Jewelry – 2003 TCO Online Round 4

https://community.topcoder.com/stat?c=problem_statement&pm=1166&rd=4705

You have been given a list of jewelry items that must be split amongst two people: Frank and Bob. Frank likes very expensive jewelry. Bob doesn't care how expensive the jewelry is, as long as he gets a lot of jewelry. Based on these criteria you have devised the following policy: 1) Each piece of jewelry given to Frank must be valued greater than or equal to each piece of jewelry given to Bob. In other words, Frank's least expensive piece of jewelry must be valued greater than or equal to Bob's most expensive piece of jewelry. 2) The total value of the jewelry given to Frank must exactly equal the total value of the jewelry given to Bob. 3) There can be pieces of jewelry given to neither Bob nor Frank. 4) Frank and Bob must each get at least 1 piece of jewelry. Given the value of each piece, you will determine the number of different ways you can allocate the jewelry to Bob and Frank following the above policy. For example: values = {1,2,5,3,4,5} Valid allocations are: Bob Frank 1,2 3 1,3 4 1,4 5 (first 5) 1,4 5 (second 5) 2,3 5 (first 5) 2,3 5 (second 5) 5 (first 5) 5 (second 5) 5 (second 5) 5 (first 5) 1,2,3,4 5,5 Note that each '5' is a different piece of jewelry and needs to be accounted for separately. There are 9 legal ways of allocating the jewelry to Bob and Frank given the policy, so your method would return 9.   Definition      Class: Jewelry Method: howMany Parameters: int[] Returns: long Method signature: long howMany(int[] values) (be sure your method is public)        Constraints - values will contain between 2 and 30 elements inclusive. - Each element of values will be between 1 and 1000 inclusive.   Examples 0)      {1,2,5,3,4,5} Returns: 9 From above. 1)      {1000,1000,1000,1000,1000, 1000,1000,1000,1000,1000, 1000,1000,1000,1000,1000, 1000,1000,1000,1000,1000, 1000,1000,1000,1000,1000, 1000,1000,1000,1000,1000} Returns: 18252025766940 2)      {1,2,3,4,5} Returns: 4 Valid allocations: Bob Frank 1,2 3 2,3 5 1,3 4 1,4 5 3)      {7,7,8,9,10,11,1,2,2,3,4,5,6} Returns: 607 4)      {123,217,661,678,796,964,54,111,417,526,917,923} Returns: 0 This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2010, TopCoder, Inc. All rights reserved.

https://blog.csdn.net/makeway123/article/details/45583645

给定一个数组，要把这个数组中的一部分数给A，另一部分给B，使A和B的和相同但是A中最小的数不小于B中最大的数。例如：

values = {1,2,5,3,4,5}

一共有9中分配方法：

 Bob       Frank
  1,2         3
  1,3         4
  1,4         5  (first 5)
  1,4         5  (second 5)
  2,3         5  (first 5)
  2,3         5  (second 5)
   5  (first 5) 5  (second 5)
   5  (second 5) 5  (first 5)
1,2,3,4       5,5

需要注意的是不同位置的相等数字需要单独计算。数组大小最大30，每个数最大1000。

https://github.com/livingstonese/topcoder/blob/master/src/main/java/tco03/onlineround4/Jewelry.java

  // Maximum sum that can be formed.

  int MAX_SUM = 30000;

  // Maximum no of jewelery items.

  int MAX_ELEMENT = 30;

  // cache for the binmoial coefficient n choose k.

  long[][] comb = new long[MAX_ELEMENT + 1][MAX_ELEMENT + 1];

  public long howMany(int[] values) {

    Arrays.sort(values);

    long ret = 0;

    int groupSize = 1;

    for (int i = 0; i < values.length - 1; i += groupSize) {

      long[] waysBelow = getSumWays(Arrays.copyOfRange(values, 0, i));

      groupSize = 1;

      for (int j = i + 1; j < values.length; j++) {

        if (values[j] != values[i])

          break;

        groupSize++;

      }

      for (int g = 0; g < groupSize; g++) {

        long[] waysAbove = getSumWays(Arrays.copyOfRange(values, i + g + 1, values.length));

        for (int s = values[i] * (g + 1); s <= MAX_SUM; s++) {

          ret += (comb(groupSize, g + 1) * waysBelow[s - values[i]*(g+1)] * waysAbove[s]);

        }

      }

    }

    return ret;

  }

  long[] getSumWays(int[] values) {

    int maxSum = sum(values);

    long[] ways = new long[MAX_SUM + 1];

    ways[0] = 1;

    for (int i = 0; i < values.length; i++) {

      for (int j = maxSum; j >= values[i]; j--) {

        ways[j] += ways[j - values[i]];

      }

    }

    return ways;

  }

  int sum(int[] values) {

    int sum = 0;

    for (int i = 0; i < values.length; i++) {

      sum += values[i];

    }

    return sum;

  }

  long comb(int n, int k) {

    if (comb[n][k] > 0)

      return comb[n][k];

    if (k == 0 || n == k)

      return 1;

    comb[n][k] = comb(n - 1, k - 1) + comb(n - 1, k);

    return comb[n][k];

  }

https://github.com/emiliot/topcoder/blob/master/TCO%202003%20online%20round%204/Jewelry.cpp