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 2^31.

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：要求把一个集合分成两个不相交的集合，使得这两个集合的元素个数相差最小的前提下，两个集合的总和之差最大。
分析：想要两个元素的个数相差最小，应当平均分；在集合元素的数量差最小的情况下，两个集合的元素之和的差最大，则前一个集合应该是最小的一半元素，后一个集合是剩下的一半元素。如果总元素数量为偶数，为了让差值最大，应该后一半元素数量多1。

因此，当总元素数量为奇数，则元素的数量差为1，否则为0.计算和的差时，先计算所有元素之和，再将数组排序后，减去前一半元素数量和的两倍。

#include<algorithm>
#include <iostream>
#include  <cstdlib>
#include  <cstring>
#include   <string>
#include   <vector>
#include   <cstdio>
#include    <queue>
#include    <stack>
#include    <ctime>
#include    <cmath>
#include      <map>
#include      <set>
#define INF 0xffffffff
#define db1(x) cout<<#x<<"="<<(x)<<endl
#define db2(x,y) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<endl
#define db3(x,y,z) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<endl
#define db4(x,y,z,r) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<endl
#define db5(x,y,z,r,w) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<", "<<#w<<"="<<(w)<<endl
using namespace std;

int main(void)
{
    #ifdef test
    freopen("in.txt","r",stdin);
    //freopen("in.txt","w",stdout);
    clock_t start=clock();
    #endif //test

    int n,ans=0;scanf("%d",&n);
    int num[n+5]={0};
    for(int i=0;i<n;++i)
        scanf("%d",&num[i]),ans+=num[i];
    sort(num,num+n);

    printf("%d ",n&1?1:0);
    for(int i=0;i<n/2;++i)
        ans-=num[i]*2;
    printf("%d\n",ans);

    #ifdef test
    clockid_t end=clock();
    double endtime=(double)(end-start)/CLOCKS_PER_SEC;
    printf("\n\n\n\n\n");
    cout<<"Total time:"<<endtime<<"s"<<endl;        //s为单位
    cout<<"Total time:"<<endtime*1000<<"ms"<<endl;    //ms为单位
    #endif //test
    return 0;
}