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

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

int main(){
    int a[1000001];
    int n; 
    cin >> n;
    int total_sum = 0;
    for(int i = 0; i < n; i++){

        scanf("%d", &a[i]);
        total_sum += a[i];

    }
    sort(a, a + n);
    if(n % 2 == 0){
        printf("0 ");
    }else
    {
        printf("1 ");
    }
    for(int i = 0; i < n / 2; i++){

        total_sum -= a[i] * 2;

    }
    printf("%d\n", total_sum);

    return 0;
}