PAT 1113 Integer Set Partition (25)

1113 Integer Set Partition (25)（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<=10^5), 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 2^​31​​.

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

思路：

题意大概就是把一组数分成两组，然后让两组的个数差最小，数字和差最大。首先在输入时可以先把总和求出来，然后把这组数排个序，取其中的一半再求个和。|n1-n2|其实就两种结果如果总个数是奇数，那么就是1，否则为0。|S1-S2|的话就是(sum-half)-half=sum-2*half。

代码：

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>  
#include <set>
using namespace std;

int main()
{
	int n, a[100005], sum = 0, half = 0;
	cin >> n;
	for (int i = 0; i < n; i++)
	{
		cin >> a[i];
		sum += a[i];
	}
	sort(a, a + n);
	for (int i = 0; i < n / 2; i++)
		half += a[i];
	cout << n % 2 << " " << sum - 2 * half << endl;
	return 0;
}