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[], N;
 int main(){
     scanf("%d", &N);
     for(int i = ; i < N; i++){
         scanf("%d", &num[i]);
     }
     sort(num, num + N);
     int mid = N / ;
     int sum1 = , sum2 = ;
     for(int i = mid; i < N; i++){
         sum2 += num[i];
     }
     for(int i = mid - ; i >= ; i--){
         sum1 += num[i];
     }
     if(N %  == )
         printf("%d %d", , sum2 - sum1);
     else printf("%d %d", , sum2 - sum1);
     cin >> N;
     return ;
 }