1113. Integer Set Partition (25)

最新推荐文章于 2021-08-24 10:48:36 发布

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本文介绍了一种算法，用于将一组正整数等分为两个子集，使得两子集元素数量之差最小，并在此基础上使两子集元素总和之差最大化。输入包括整数的数量及具体的整数值，输出为两子集的元素数量之差及其总和之差。

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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

题目大意：

代码：

#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
bool cmp(long long a,long long b)
{
	return a>b;
}
vector<long long> v;
int main()
{
	long long i,j,n,m,k,t,S1=0,S2=0;
	scanf("%lld",&n);
	for(i=0;i<n;i++)
	{
		scanf("%lld",&m);
		v.push_back(m);
	}
	sort(v.begin(),v.end(),cmp);
	for(i=0;i<v.size();i++)
	{
		if(i<((n+1)/2))
		{
			S1+=v[i];
		}
		else
		{
			S2+=v[i];
		}
	}
	if(n%2==0)
	{
		printf("0 ");
	}
	else
	{
		printf("1 ");
	}
	printf("%lld\n",S1-S2);
	return 0;
}