1113 Integer Set Partition (25 分)

1113 Integer Set Partition (25 分)

Given a set of N (>) 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 ∣ is minimized first, and then ∣ is maximized.

Input Specification:

Each input file contains one test case. For each case, the first line gives an integer N (2), 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: ∣ and ∣, 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

排个序，判个奇偶

 1 #include <bits/stdc++.h>
 2
 3 using namespace std;
 4 int n, an[200005];
 5 int main(){
 6     cin >> n;
 7     for(int i = 0; i < n; i++){
 8         cin >> an[i];
 9     }
10     sort(an, an+n);
11     int ans = 0, cnt = 0;
12     for(int i = 0; i < n/2; i++){
13         ans += an[i];
14     }
15     for(int i = n/2; i < n; i++){
16         cnt += an[i];
17     }
18     cout << (n&1) <<" "<<cnt-ans<<endl;
19     return 0;
20 }

原文地址：https://www.cnblogs.com/zllwxm123/p/11334541.html