【PAT】【Advanced Level】1113. Integer Set Partition (25)

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本文介绍了一个算法问题，即如何将一组正整数合理地划分为两个子集，使得这两个子集的元素数量之差最小，并在此基础上使元素总和之差最大化。提供了具体的输入输出规范与示例，以及一种基于排序和两端向中间累加的方法。

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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 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

原题链接：

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

思路：

排序，从两边向中间累加前后之差，最后针对中间的元素，使累加的绝对值尽可能的大

对于个数差，奇数为1，偶数为0

CODE：

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#define N 100010
using namespace std;
int num[N];
int main()
{
	int n;
	scanf("%d",&n);
	for (int i=0;i<n;i++)
	{
		scanf("%d",&num[i]);
	}
	sort(num,num+n);
	int c=0;
	for (int i=0,j=n-1;i<=j;i++,j--)
	{
		c+=(num[j]-num[i]);
	}
	if (n%2==1)
	{
		if (num[n/2]>=0)
		{
			if (c>=0)
			{
				c+=num[n/2];
			}
			else
			{
				c-=num[n/2];
			}
		}
		else
		{
			if (c<0)
			{
				c+=num[n/2];
			}
			else
			{
				c-=num[n/2];
			}
		}
	}	
	if (n%2==1)
	{
		cout<<1<<" ";
	}
	else
	{
		cout<<0<<" ";
	}
	if (c<0)
	{
		cout<<c*-1;
	}
	else
	{
		cout<<c;
	}
	return 0;
}