單詞積累

disjoint 打散 拆開

思路

分成兩個(gè)集合，數(shù)量之差最小，總和之差盡可能大。排序后平分即可

代碼

#include <bits/stdc++.h>
using namespace std;

const int maxn = 100003;
int n;
int num[maxn];

int main() {
    cin>>n;
    for (int i = 0; i < n; i++) {
        cin>>num[i];
    }
    sort(num, num + n);
    if (n % 2 == 1) cout<<1<<" ";
    else cout<<0<<" ";
    int ans1 = 0, ans2 = 0;
    for (int i = 0; i < n / 2; i++) {
        ans1 += num[i];
    }
    for (int i = n / 2; i < n; i++) {
        ans2 += num[i];
    }
    cout<<ans2-ans1<<endl;
}