L2-017 人以群分 (25分)

L2-017 人以群分 (25分)

社交网络中我们给每个人定义了一个“活跃度”，现希望根据这个指标把人群分为两大类，即外向型（outgoing，即活跃度高的）和内向型（introverted，即活跃度低的）。要求两类人群的规模尽可能接近，而他们的总活跃度差距尽可能拉开。

输入格式：
输入第一行给出一个正整数N（2≤N≤10
​5
​​ ）。随后一行给出N个正整数，分别是每个人的活跃度，其间以空格分隔。题目保证这些数字以及它们的和都不会超过2
​31
​​ 。

输出格式：
按下列格式输出：

Outgoing #: N1
Introverted #: N2
Diff = N3
其中N1是外向型人的个数；N2是内向型人的个数；N3是两群人总活跃度之差的绝对值。

输入样例1：

10
23 8 10 99 46 2333 46 1 666 555

输出样例1：

Outgoing #: 5
Introverted #: 5
Diff = 3611

输入样例2：

13
110 79 218 69 3721 100 29 135 2 6 13 5188 85

输出样例2：

Outgoing #: 7
Introverted #: 6
Diff = 9359

代码：

//输入n个数，排序；偶数个：奇数个；
#include
#include 
#include 
using namespace std;
bool cmp(int a, int b) {
     
	return a > b;
}
int main() {
     
	int n;
	cin >> n;
	int a[100000];
	for (int i = 0; i < n; i++) cin >> a[i];
	sort(a, a + n, cmp);
	if (n % 2 == 0) {
     
		int sum1=0, sum2 = 0;
		for (int i = 0; i < n / 2; i++) sum1 += a[i];
		for (int i = n / 2; i < n; i++) sum2 += a[i];
		int diff = sum1 - sum2;
		cout << "Outgoing #: " << n / 2 << endl;
		cout << "Introverted #: " << n / 2 << endl;
		cout << "Diff = " << diff << endl;
	}
	else {
     
		int sum1 = 0, sum2 = 0;
		for (int i = 0; i < n / 2+1; i++) sum1 += a[i];
		for (int i = n / 2+1; i < n; i++) sum2 += a[i];
		int diff = sum1 - sum2;
		cout << "Outgoing #: " << n / 2+1 << endl;
		cout << "Introverted #: " << n / 2 << endl;
		cout << "Diff = " << diff << endl;
	}
}