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 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: ∣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
1 #include <stdio.h>
2 #include <algorithm>
3 using namespace std;
4 const int maxn=100010;
5 int seq[maxn];
6 int total[maxn];
7 int main(){
8 int n;
9 int sum=0;
10 scanf("%d",&n);
11 for(int i=0;i<n;i++){
12 scanf("%d",&seq[i]);
13 }
14 sort(seq,seq+n);
15 for(int i=0;i<n;i++){
16 sum+=seq[i];
17 total[i]=sum;
18 }
19 printf("%d %d",n%2,total[n-1]-2*total[n/2-1]);
20 }
注意点：计算差的时候由于总和算了前面部分，要多减一次前半部分。感觉直接算和然后相减也不会超时
原文地址：https://www.cnblogs.com/tccbj/p/10438206.html