A1113. Integer Set Partition

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<cstdio>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
int num[100001], N;
int main(){
    scanf("%d", &N);
    for(int i = 0; i < N; i++){
        scanf("%d", &num[i]);
    }
    sort(num, num + N);
    int mid = N / 2;
    int sum1 = 0, sum2 = 0;
    for(int i = mid; i < N; i++){
        sum2 += num[i];
    }
    for(int i = mid - 1; i >= 0; i--){
        sum1 += num[i];
    }
    if(N % 2 == 0)
        printf("%d %d", 0, sum2 - sum1);
    else printf("%d %d", 1, sum2 - sum1);
    cin >> N;
    return 0;
}