PAT 1113 Integer Set Partition（25 分）

最新推荐文章于 2022-11-15 17:07:48 发布

chenxli23

最新推荐文章于 2022-11-15 17:07:48 发布

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本文链接：https://blog.youkuaiyun.com/Hickey_Chen/article/details/81517926

本文介绍了一种算法，用于将一组正整数分为两个子集，使得这两个子集的元素数量之差最小，并在此基础上使子集元素总和之差最大化。通过排序并按顺序分配元素到两个子集中实现这一目标。

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Given a set of N (>1) positive integers, you are supposed to partition them into two disjoint sets A1 and A2 of n1 and n2 numbers, respectively. Let S1 and S2 denote the sums of all the numbers in A1 and A2, respectively. You are supposed to make the partition so that ∣n1−n2∣ is minimized first, and then ∣S1−S2∣ is maximized.

Input Specification:

Each input file contains one test case. For each case, the first line gives an integer N (2≤N≤105), and then N positive integers follow in the next line, separated by spaces. It is guaranteed that all the integers and their sum are less than 231.

Output Specification:

For each case, print in a line two numbers: ∣n1−n2∣ and ∣S1−S2∣, separated by exactly one space.

Sample Input 1:

10
23 8 10 99 46 2333 46 1 666 555

Sample Output 1:

0 3611

Sample Input 2:

13
110 79 218 69 3721 100 29 135 2 6 13 5188 85

Sample Output 2:

1 9359

思路：要使两个集合的差最大，把数排序后前n/2个数是一个集合，剩下的是一个集合。

程序：

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{
  int n;
  scanf("%d",&n);
  vector<int> v;
  for(int i = 0; i < n; i++)
  {
    int val;
    scanf("%d",&val);
    v.push_back(val);
  }
  sort(v.begin(),v.end());
  int sum1 = 0, sum2 = 0;
  for(int i = 0; i < n / 2; i++)
  {
    sum1 += v[i];
  }
  for(int i = n / 2; i < n; i++)
  {
    sum2 += v[i];
  }
  if(n % 2 == 0)
  {
    printf("0 %d\n",sum2-sum1);
  }
  else
  {
    printf("1 %d\n",sum2-sum1);
  }
  return 0;
}