人以群分

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≤105)。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231。

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
#include 
using namespace std;
typedef long long LL;
const int N = 1e5 + 10, M = 10010;
int n;
int a[N];

int main() {
    scanf("%d", &n);
    LL res = 0, sum = 0;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
    }

    sort(a + 1, a + n + 1);
    for (int i = 1; i <= n; i++) {
        if (i <= n / 2)res += (LL)a[i];
        else sum += a[i];
    }
    cout << "Outgoing #: " << n - (n / 2) << endl;
    cout << "Introverted #: " << n / 2 << endl;
    cout << "Diff = " << sum - res;
    return 0;
}

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