1113 Integer Set Partition (25 point(s))

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 A​1​​ and A​2​​ of n​1​​and n​2​​ numbers, respectively. Let S​1​​ and S​2​​ denote the sums of all the numbers in A​1​​ and A​2​​, respectively. You are supposed to make the partition so that ∣n​1​​−n​2​​∣ is minimized first, and then ∣S​1​​−S​2​​∣ is maximized.

Input Specification:

Each input file contains one test case. For each case, the first line gives an integer N (2≤N≤10​5​​), 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 2​31​​.

Output Specification:

For each case, print in a line two numbers: ∣n​1​​−n​2​​∣ and ∣S​1​​−S​2​​∣, 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;
}