PAT甲级——1113 Integer Set Partition (排序)

最新推荐文章于 2022-03-14 16:36:11 发布

Little Panda

最新推荐文章于 2022-03-14 16:36:11 发布

阅读量348

点赞数

分类专栏： PAT (Advanced Level) Practice

本文链接：https://blog.youkuaiyun.com/weixin_44385565/article/details/89949463

版权

PAT (Advanced Level) Practice 专栏收录该内容

50 篇文章

订阅专栏

博客围绕整数集分区问题展开，给定N个正整数，需将其分成两个不相交集合A1和A2，使|n1 - n2|最小且|S1 - S2|最大。输入为N及N个正整数，输出为|n1 - n2|和|S1 - S2|的值。思路是将数组排序，前半部分为S2，后半部分为S1。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 A1and 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

题目大意：将N个整数分成两个集合S1、S2（S1和S2分别为当前集合的元素之和），元素个数分别为n1、n2；找出使得|n1-n2|最小、同时|S1-S2|最大的分类操作，并输出|n1-n2|和|S1-S2|的值。

思路：数组从小到大排序，前一半是S2，后一半是S1，送分题~

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
	int N;
	scanf("%d",&N);
	vector<int> v(N);
	for(int i=0;i<N;i++){
		scanf("%d",&v[i]);
	}
	sort(v.begin(),v.end());
	int n2=N/2,n1=N-n2,S1=0,S2=0;
	for(int i=0;i<n2;i++)
		S2+=v[i];
	for(int i=n2;i<N;i++)
		S1+=v[i];
	printf("%d %d",n1-n2,S1-S2);
	return 0;
}