文章目录

• A1113 Integer Set Partition (25 point(s))
• Input Specification:
• Output Specification:
• Sample Input 1:
• Sample Output 1:
• Sample Input 2:
• Sample Output 2:

A1113 Integer Set Partition (25 point(s))

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 <stdio.h>
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main(){
    int n,sum1=0,sum=0;
    scanf("%d",&n);
    vector<int> vi(n);
    for(int i=0;i<n;i++)    scanf("%d",&vi[i]);
    sort(vi.begin(),vi.end());
    for(int i=0;i<n/2;i++)   sum1+=vi[i];
    for(int i=0;i<n;i++) sum+=vi[i];
    printf("%d %d\n",n-n/2-n/2,sum-sum1-sum1);
    return 0;
}