PAT甲级真题 1113 Integer Set Partition (25分) C++实现 (简单划分数组)

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#c++ #pat #划分 #分组

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本文介绍了一种针对一组正整数的最优划分算法，旨在将这些整数划分为两个子集，使得两子集元素数量之差最小，且两子集元素总和之差最大化。算法首先对整数进行排序，然后将前半部分较小的数放入一个子集，后半部分较大的数放入另一个子集，最后输出元素数量差和元素总和差。

题目

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

思路

要求划分后个数差值最小，那么只能是尽可能平分后，相差1或0个。

差值最大就是将数组排序后，较大的一部分放进前一个分组，后一部分放入后面的分组，求二者和的差值即可。

代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main()
{
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i=0; i<n; i++){
        cin >> a[i];
    }
    sort(a.begin(), a.end());
    
    int sum1 = 0;
    for (int i=0; i<n/2; i++){
        sum1 += a[i];
    }
    int sum2 = 0;
    for (int i=n/2; i<n; i++){
        sum2 += a[i];
    }
    cout << n % 2 << " " << sum2 - sum1;
    return 0;
}