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;
}