1113 Integer Set Partition （25 分）集合分割

1113 Integer Set Partition （25 分）

Given a set of NNN (>1> 1>1) positive integers, you are supposed to partition them into two disjoint sets A1A_1A​1​​ and A2A_2A​2​​ of n1n_1n​1​​ and n2n_2n​2​​ numbers, respectively. Let S1S_1S​1​​ and S2S_2S​2​​ denote the sums of all the numbers in A1A_1A​1​​ and A2A_2A​2​​, respectively. You are supposed to make the partition so that ∣n1−n2∣|n_1 - n_2|∣n​1​​−n​2​​∣ is minimized first, and then ∣S1−S2∣|S_1 - S_2|∣S​1​​−S​2​​∣ is maximized.

Input Specification:

Each input file contains one test case. For each case, the first line gives an integer NNN (2≤N≤1052 le N le 10^52≤N≤10​5​​), and then NNN positive integers follow in the next line, separated by spaces. It is guaranteed that all the integers and their sum are less than 2312^{31}2​31​​.

Output Specification:

For each case, print in a line two numbers: ∣n1−n2∣|n_1 - n_2|∣n​1​​−n​2​​∣ and ∣S1−S2∣|S_1 - S_2|∣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

思路：
　　由题目可知，当n是偶数时，n1=n2；当n是奇数时，n2-n1=1，计算abs(sum1-sum2),首先对整个集合进行排序，
然后计算即可。

#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
#include<string>
#include<map>
#include<set>
using namespace std;



int main()
{
    int n;
    cin>>n;
    int a[n+1];
    for(int i=1;i<n+1;i++)
        cin>>a[i];
    sort(a+1,a+n+1);

    if(n%2==0)
    {
        long long int sum=0;
        for(int i=1;i<=n/2;i++)
        {
            sum+=a[n-i+1]-a[i];
        }
        cout<<"0 "<<sum;
    }
    else
    {
        long long int sum=0;
        for(int i=1;i<=n/2;i++)
        {
            sum+=a[n-i+1]-a[i];
        }
        sum+=a[n/2+1];
        cout<<"1 "<<sum;
    }
    return 0;
}