1113 Integer Set Partition

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

 

题意：

　　将一个给定的数组分成两部分，使两部分元素的个数之差最小，总和之差最大。

思路：

　　将数组排序，从中间开始划分，前一半的和最小，后一半的和最大，满足题意。

Code：

#include <bits/stdc++.h>

using namespace std;

int sum(vector<int> v, int s, int e) {
    int sum = 0;
    for (int i = s; i <= e; ++i) {
        sum += v[i];
    }
    return sum;
}

int main() {
    int n;
    cin >> n;
    vector<int> v(n);
    for (int i = 0; i < n; ++i) cin >> v[i];
    sort(v.begin(), v.end());
    int mid = n / 2;
    int preSum = sum(v, 0, mid - 1);
    int postSum = sum(v, mid, n - 1);
    if (v.size() % 2 == 0) {
        cout << 0 << " " << postSum - preSum << endl;
    } else {
        cout << 1 << " " << postSum - preSum << endl;
    }
    return 0;
}