1113 Integer Set Partition (25 分)

1113 Integer Set Partition (25 分)

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

题目大意：
给定一个集合，要求分为两个集合，使这两个集合的元素个数相差最小，元素和相差最大

AC代码：

#include
using namespace std;
int main()
{
	int num,s1=0,s2=0;
	cin>>num;
	vector<int> v(num);
	for(int i=0;i<num;i++)
	scanf("%d",&v[i]);
	sort(v.begin(),v.end());
	for(int i=0;i<num/2;i++)
	s1+=v[i];
	for(int i=num/2;i<num;i++)
	s2+=v[i];
	printf("%d %d",num%2,s2-s1);
}