PAT 1113 Integer Set Partition

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最新推荐文章于 2025-06-13 19:04:42 发布

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这篇博客讲解了一种算法，如何将一组正整数划分为两个集合，使得两个集合中数字个数差的绝对值最小，同时两集合和的绝对差值最大。通过实例演示和代码实现，适合考研准备者理解并应用于算法问题解决。

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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

总结：正数划分，很简单的一道题目，把数字分成两个集合，两个集合中数字的个数最小，集合的和之查最大，直接将n/2，就是第一个集合的数字个数，将数组排好序后，从小到大将前n/2个数字相加，值就是第一个集合的和，剩下数字的和和数字个数就是集合②的信息

代码：

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

int main(){
    int n;
    scanf("%d",&n);
    int a[n];
    
    for(int i=0;i<n;i++)    scanf("%d",&a[i]);
    sort(a,a+n);
    int s1=0,s2=0,n1=n/2,n2=n-n1;
    for(int i=0;i<n1;i++)   s1+=a[i];
    for(int i=n1;i<n;i++)   s2+=a[i];
    printf("%d %d\n",n2-n1,s2-s1);
    return 0;
}

好好学习，天天向上！

我要考研！