1113 Integer Set Partition（25 分）

1113 Integer Set Partition（25 分）

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

代码:

#include
using namespace std;
const int maxn = 1e5+10;
int digit[maxn];
int main()
{
	int n;
	scanf("%d",&n);
	for(int i = 1; i <=n ; i++)
		scanf("%d",&digit[i]);
	sort(digit+1,digit+1+n);//从小到大排序之后		
	int sum1 =0 ,sum2 = 0;
	for(int i = 1; i <= n/2;i++)
			sum1 += digit[i];
	for(int i = n / 2 + 1; i<=n;i++) 
			sum2 += digit[i];
	if(n % 2 == 1)	
		printf("1 %d\n",sum2-sum1);	
	else 
		printf("0 %d\n",sum2-sum1);
	return 0;
}
//这题目的意思就是说 要是他们的个数相差是尽可能的小
//并且要满足的是他们的和之间是尽可能的大