Given a set of N (>1) positive integers, you are supposed to partition them into two disjoint sets A​1​​ and A​2​​ of n​1​​ and n​2​​numbers, respectively. Let S​1​​ and S​2​​ denote the sums of all the numbers in A​1​​ and A​2​​, respectively. You are supposed to make the partition so that ∣n​1​​−n​2​​∣ is minimized first, and then ∣S​1​​−S​2​​∣ is maximized.

Input Specification:

Each input file contains one test case. For each case, the first line gives an integer N (2≤N≤10​5​​), 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 2​31​​.

Output Specification:

For each case, print in a line two numbers: ∣n​1​​−n​2​​∣ and ∣S​1​​−S​2​​∣, 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

題意：把一組數分成兩個集合，在N2-N1最小的前提下，保證S2-S1最大。

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
	int n;
	cin >> n;
	int sum = 0;//集合之和
	int cnt;
	int *arry = new int[n];
	for (int i = 0; i < n; i++)
	{
		cin >> arry[i];
		sum += arry[i];
	}
	sort(arry, arry + n);
	int sum1 = 0;//集合一半之和
	for (int i = 0; i < n / 2; i++)
		sum1 += arry[i];
	cout << n % 2 << " " << sum - 2*sum1 << endl;
	return 0;
}