1113 Integer Set Partition (25 分)-优快云博客

本文链接：https://blog.youkuaiyun.com/Tropine/article/details/120511292

该博客讨论了一种算法问题，即如何将一组正整数分成两组，使得两组元素数量之差最小，并且这两组的和之差最大化。输入包含一组不超过10^5个的正整数，要求首先最小化数量差，然后最大化和的差。程序通过排序整数并等分（如果数量为偶数）或接近等分（如果数量为奇数）来实现这一目标。

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1113 Integer Set Partition (25 分)

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

Analysis

简单题，无坑点无槽点

Code(C++)

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

int main()
{
    int num;
    cin >> num;
    int a[num];
    for(int i=0; i<num; i++)
        cin >> a[i];
    
    sort(a, a+num);
    
    int n1 = num / 2;
    int n2 = num - n1;
    int sum1 = 0, sum2 = 0;
    for(int i=0; i<n1; i++)
        sum1 += a[i];
    for(int i=n1; i<num; i++)
        sum2 += a[i];
    
    cout << n2-n1 << " " << sum2 - sum1 << endl;
}