PAT A1113 Integer Set Partition (25point(s))

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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

题意

给定一个数集,将该数集拆分为两个互不相交的子集,使它们的集合大小差最小,然后使它们的和差最大

解答

显然集合大小差为0或1,当集合大小为偶数时,子集大小差为0,当集合大小为偶数时,子集大小差为1。要使和差最大,则要对集合进行从小到大排序,前半部分给S1,后半部分给S2,并使S2的大小不比S1小,即SIZE(S2)-SIZE(1)等于0或1

#include<iostream>
#include<algorithm>
#include<vector>

using namespace std;

int main()
{
	int n, s1 = 0, s2 = 0;
	cin >> n;
	vector<int> arr(n, 0);
	for (int i = 0; i < n; i++)
		cin >> arr[i];
	sort(arr.begin(), arr.end());
	int mid = n / 2;
	for (int i = 0; i < mid; i++)
		s1 += arr[i];
	for (int i = mid; i < n; i++)
		s2 += arr[i];
	if (n % 2 == 0) cout << "0 " << s2 - s1;
	else cout << "1 " << s2 - s1;
	return 0;
}