【PAT甲级 - C++题解】1113 Integer Set Partition

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专栏地址：PAT题解集合
原题地址：题目详情 - 1113 Integer Set Partition (pintia.cn)
中文翻译：整数集合划分
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1113 Integer Set Partition

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

题意

给定 n 个数，将这 n 个数划分成两个集合 A1 和 A2 ，其中 A1 包含 n1 个数且这些数之和为 s1 ，A2 包含 n2 个数且这些数之和为 s2 。

现在需要我们使 |n2-n1| 的值尽可能的小，且 |s2-s1| 的值尽可能的大。

思路

直接上结论，我们只要将数组先排个序，然后尽可能均匀的分成两部分，所以 |n2-n1| 的值只可能为 0 或 1 ，且左半部分的值都是小于右半部分的，这样就能使 |s2-s1| 的值最大。

代码

#include
using namespace std;

const int N = 100010;
int w[N];
int n;

int main()
{
    cin >> n;
    for (int i = 0; i < n; i++)    cin >> w[i];

    //进行排序
    sort(w, w + n);

    //计算两部分数值的和
    int s1 = 0, s2 = 0;
    for (int i = 0; i < n / 2; i++)  s1 += w[i];
    for (int i = n / 2; i < n; i++)  s2 += w[i];

    //输出结果
    cout << n % 2 << " " << s2 - s1 << endl;

    return 0;
}