1113 Integer Set Partition (25分)

最新推荐文章于 2025-09-28 10:15:29 发布

原创最新推荐文章于 2025-09-28 10:15:29 发布 · 97 阅读

0 ·

CC 4.0 BY-SA版权

PAT 专栏收录该内容

126 篇文章

订阅专栏

本文探讨了如何将一组正整数N个元素分为两个不相交子集A1和A2，使得两子集的元素数量之差的绝对值最小，且在此基础上两子集元素和之差的绝对值最大化。输入包括整数N及N个正整数值，输出为两子集元素数量差和元素和差。

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

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

int main(){
    int a[1000001];
    int n; 
    cin >> n;
    int total_sum = 0;
    for(int i = 0; i < n; i++){

        scanf("%d", &a[i]);
        total_sum += a[i];

    }
    sort(a, a + n);
    if(n % 2 == 0){
        printf("0 ");
    }else
    {
        printf("1 ");
    }
    for(int i = 0; i < n / 2; i++){

        total_sum -= a[i] * 2;

    }
    printf("%d\n", total_sum);

    return 0;
}