Integer Set Partition

最新推荐文章于 2022-11-15 17:07:48 发布

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本文介绍了一种将正整数集合划分为两个子集的方法，目标是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的绝对值。通过排序并按顺序分配元素到两个子集的方式实现此目标。

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

一道水题

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int a[100000];
int main() {
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++)
		scanf("%d",&a[i]);
	sort(a,a+n);
	int sum1=0;
	for(int i=0;i<n/2;i++)
		sum1+=a[i];
	int sum2=0;
	for(int i=n/2;i<n;i++)
		sum2+=a[i];
	printf("%d ",n%2);
	printf("%d",sum2-sum1);
	return 0;
}