PAT--1113 Integer Set Partition (25 分)

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本文探讨了如何将一组正整数划分成两个不相交子集，以最小化两子集元素数量之差，并在此基础上最大化两子集元素总和之差。通过排序和合理分配元素，实现算法的高效求解。

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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

代码

#include <iostream>
#include <bits/stdc++.h>

using namespace std;


int main()
{
    int n;
    int sum = 0;
    int counts = 0;
    cin>>n;
    int a[n];
    for(int i=0;i<n;i++){
        cin>>a[i];
        counts +=a[i];
    }
    sort(a,a+n);
    int flag;
    if(n%2==0){
        flag = 0;
    }else{
        flag = 1;
    }
    for(int i=0;i<n/2;i++){
        sum +=a[i];
    }

    cout<<flag<<" "<<counts-2*sum<<endl;
    return 0;
}
分析：

刚开始以为是在不改变集合顺序的情况下求，想到用求出每个部分再求最大，但是没有思路。在借鉴其他人之后，才知道可以改变顺序，那就很简单。