1113. Integer Set Partition (25)

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1113. Integer Set Partition (25)

时间限制

150 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Given a set of N (> 1) positive integers, you are supposed to partition them into two disjoint sets A₁ and A₂ of n₁ and n₂numbers, 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

题意：给出n个数据，要求把其分为两组，使得两组的元素个数之差最小 并且 两组值总和之差最大

分析：_(:з」∠)_ 是因为前面那题太坑了所以这题这么水吗

         个数差最小，很明显一半一半，偶数差就是0了，奇数差就是1

         值差最大，排序一下，后半段减前半段就行了

         虽然交上去是AC，但是我总感觉真实的题意被我略过了……

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        （在想集合set要不要考虑重复的数字）

代码：

#include <iostream>       
#include <vector>         
#include <algorithm>
using namespace std;
vector<int> num;
int main ()
{
	int n=0,sum=0,s=0;
	cin >> n;
	for(int i=0;i<n;i++)
	{
		int tmp;
		cin>>tmp;
		num.push_back(tmp);
		s+=tmp;
	}
	sort(num.begin(),num.end());
	for(int i=0;i<n/2;i++)
		sum+=num[i];
	cout<<n%2<<" "<<s-2*sum<<endl;
}