1113. Integer Set Partition (25)

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:

1023 8 10 99 46 2333 46 1 666 555

Sample Output 1:

0 3611

Sample Input 2:

13110 79 218 69 3721 100 29 135 2 6 13 5188 85

Sample Output 2:

1 9359

偶数平分，奇数后面比前面多1

#include<cstdio>#include<cmath>#include<algorithm>#include<iostream>#include<cstring>#include<queue>#include<vector>#include<set>#include<map>#include<stack>using namespace std;int main(){int k;cin>>k;int a[100001];for(int i=0;i<k;i++){scanf("%d",&a[i]); }sort(a,a+k);int n2=(k+1)/2;int n1=k/2;int sum=0;for(int j=k-1;j>=k-n2;j--){sum+=a[j];}for(int i=0;i<n1;i++) sum-=a[i];cout<<n2-n1<<" "<<sum; return 0;}