1113. Integer Set Partition (25)

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本文介绍了一种算法，用于将一组正整数等分为两个子集，使得两子集元素数量之差最小，并在此基础上最大化两子集元素总和之差的绝对值。输入为一组正整数，输出为两子集元素数量之差及元素总和之差。

Given a set of N (> 1) positive integers, you are supposed to partition them into two disjoint sets A₁ and A₂ of n₁ and n₂numbers, 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

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;

int main()
{
    int n,sum1=0,sum2=0;
    cin>>n;
    vector<int> num(n);
    for(int i=0;i<n;i++)
    {
        cin>>num[i];
        sum1=sum1+num[i];
    }
    sort(num.begin(),num.end());
    for(int i=0;i<n/2;i++)
    {
        sum2=sum2+num[i];
    }
    cout<<n%2<<" "<<sum1-2*sum2<<endl;
    return 0;
}