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 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: ∣ 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

将一个数分为2个集合，求如何分，能够使个数差最小，总和差最大。

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main() {
    int n, sum = 0, halfsum = 0;
    scanf("%d", &n);
    vector<int> v(n);
    for(int i = 0; i < n; i++) {
        scanf("%d", &v[i]);
        sum += v[i];
    }
    sort(v.begin(), v.end());
    for(int i = 0; i < n / 2; i++)
        halfsum += v[i];
    printf("%d %d", n % 2, sum - 2 * halfsum);
    return 0;
}