1113. Integer Set Partition (25)

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本文介绍了一个算法问题：如何将一个正整数集合划分为两个不相交子集，使得两个子集的元素数量之差最小，并在此基础上使两子集元素和之差最大化。输入包含一组正整数，输出为子集元素数量差及元素和差。

1113. Integer Set Partition (25)
时间限制 150 ms 内存限制 65536 kB 代码长度限制 16000 B
判题程序 Standard 作者 CHEN, Yue

Given a set of N (> 1) positive integers, you are supposed to partition them into two disjoint sets A1 and A2 of n1 and n2 numbers, 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

#define _CRT_SECURE_NO_WARNINGS
#include <algorithm>
#include <iostream>

using namespace std;
const int MaxN = 100010;
int Data[MaxN];

int main() {
#ifdef _DEBUG
    freopen("data.txt", "r+", stdin);
#endif // _DEBUG

    std::ios::sync_with_stdio(false);

    int n; cin >> n;
    int diff = 0, dn = 1;
    for (int i = 1; i <= n; ++i)cin >> Data[i]; Data[0] = 0;
    sort(Data, Data + n + 1);

    int i = 0, j = n;
    while(i < j)diff += Data[j--] - Data[i++];
    if (i == j) {
        dn = 0;
        diff -= Data[i];
    }

    cout << dn << " " << diff;

    return 0;
}