1113 Integer Set Partition (25 point(s))

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本文介绍了一个整数集合划分的问题，给定一组正整数，目标是将其划分为两个不相交的子集，使得两子集元素数量之差绝对值最小，且两子集元素总和之差绝对值最大化。通过一种简单的贪心策略实现这一目标，首先对整数进行排序，然后将前一半元素分配给一个子集，后一半分配给另一个子集。

1113 Integer Set Partition (25 point(s))

Given a set of N (>1) positive integers, you are supposed to partition them into two disjoint sets A1 and A2 of n1and 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 Npositive 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

简单的贪心策略。

#include<iostream>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
ll a[100007];
int main(void){
	int N;scanf("%d",&N);
	for(int i=0;i<N;i++) scanf("%lld",&a[i]);
	sort(a,a+N);
	ll sum = 0;
	for(int i=0;i<N;i++){
		if(i<N/2) sum=sum-a[i];
		else sum+=a[i];
	}
	if(N%2==0) printf("0 ");
	else printf("1 ");
	printf("%lld\n",sum);
	return 0;
}