1113. Integer Set Partition (25)-优快云博客

本文探讨了如何将一组正整数N分为两个不相交的子集A1和A2，使得两集合中元素数量之差最小，同时两集合元素总和之差最大化。通过排序和累积求和的方法，实现了一个高效的时间复杂度为O(NlogN)的算法。

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:

23 8 10 99 46 2333 46 1 666 555

Sample Output 1:

0 3611

Sample Input 2:

110 79 218 69 3721 100 29 135 2 6 13 5188 85

Sample Output 2:

1 9359

C++:

/*
 @Date    : 2018-02-28 10:06:21
 @Author  : 酸饺子 (changzheng300@foxmail.com)
 @Link    : https://github.com/SourDumplings
 @Version : $Id$
*/

/*
https://www.patest.cn/contests/pat-a-practise/1113
 */

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

int main(int argc, char const *argv[])
{
    int N;
    scanf("%d", &N);
    int num[N];
    for (int i = 0; i != N; ++i)
        scanf("%d", &num[i]);
    sort(num, num+N);
    int n2 = N / 2;
    int S2 = 0, S1 = 0;
    S2 = accumulate(num, num+n2, 0);
    S1 = accumulate(num+n2, num+N, 0);
    if ((N & 1) == 0)
        printf("0 %d\n", S1 - S2);
    else
        printf("1 %d\n", S1 - S2);
    return 0;
}