1113. Integer Set Partition

最新推荐文章于 2022-11-15 17:07:48 发布

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最新推荐文章于 2022-11-15 17:07:48 发布

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针对一组正整数，本篇介绍一种算法实现：将其划分为两个互斥子集，目标首先是使两子集元素数量之差最小，其次是在此基础上使两子集元素总和之差尽可能大。文中提供了一个具体的C++实现方案。

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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

这个题很直白。

#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

int a[100000];
int main()
{
	int n;
	cin>>n;
	int s1=0,s2=0;
	for(int i=0;i<n;i++)
		cin>>a[i];
	sort(a,a+n);
		for(int i=0;i<n/2;i++)
			s1+=a[i];
		for(int i=n/2;i<n;i++)
			s2+=a[i];
	if(n%2==0)
		cout<<"0 ";
	else
		cout<<"1 ";
	cout<<s2-s1;
}