1113. Integer Set Partition (25)-PAT甲级真题

Given a set of N (> 1) positive integers, you are supposed to partition them into two disjoint sets A1 and A2 of n1 and n2 numbers, respectively. Let S1 and S2 denote the sums of all the numbers in A1 and A2, respectively. You are supposed to make the partition so that |n1 – n2| is minimized first, and then |S1 – S2| is maximized.

Input Specification:

Each input file contains one test case. For each case, the first line gives an integer N (2 <= N <= 105), 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 231.

Output Specification:

For each case, print in a line two numbers: |n1 – n2| and |S1 – S2|, 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


题目理解

给出N个数,分散到两个集合中,要求使得两个集合元素个数的差最小,同时两集合总和之差最大

分析

  • 将组内元素排序,计算前n/2个元素的和halfsum,n元素的总和sum-2*halfsum即为|S1- S2|
  • |n1-n2|=n%2
#include 
#include 
#include 

using namespace std;

int main() {
    int n, sum = 0, halfsum = 0;
    cin >> n;
    vector v(n);
    for (int i = 0; i < n; i++) {
        cin >> v[i];
        sum += v[i];
    }
    sort(v.begin(), v.end());
    for (int i = 0; i < n / 2; i++) {
        halfsum += v[i];
    }
    cout << n % 2 << " " << sum - 2 * halfsum << endl;
    return 0;
}

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