PAT (Advanced Level) Practice 1113 Integer Set Partition (25分)

1.题目

Given a set of N (>1) positive integers, you are supposed to partition them into two disjoint sets A​1​​ and A​2​​ of n​1​​ and n​2​​ numbers, respectively. Let S​1​​ and S​2​​ denote the sums of all the numbers in A​1​​ and A​2​​, respectively. You are supposed to make the partition so that ∣n​1​​−n​2​​∣ is minimized first, and then ∣S​1​​−S​2​​∣ 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: ∣n​1​​−n​2​​∣ and ∣S​1​​−S​2​​∣, 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

2.代码

我不敢相信就这么过了…………………………

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
	int n,sum1=0,sum2=0,count=0;
	scanf("%d", &n);
	vector<int>list(n);
	for (int i = 0; i < n; i++)
		scanf("%d", &list[i]);
	sort(list.begin(), list.end());
	if (n % 2 == 0)printf("0 ");
	else printf("1 ");
	for (auto it = list.begin(); it != list.end(); it++)
	{
		count++;
		if (count <= list.size() / 2)
			sum1 += *it;
		else
			sum2 += *it;
	}
	printf("%d", sum2 - sum1);
}