PAT L2-017 人以群分

https://pintia.cn/problem-sets/994805046380707840/problems/994805061056577536

 

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）。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

时间复杂度：$O(N * logN)$

代码：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 10;
int N;
int a[maxn];

int main() {
    scanf("%d", &N);
    for(int i = 1; i <= N; i ++)
        scanf("%d", &a[i]);

    sort(a + 1, a + 1 + N);
    int num1 = 0, num2 = 0;
    long long sum1 = 0, sum2 = 0, diff = 0;

    if(N % 2) {
        num1 = (N - 1) / 2;
        num2 = N - num1;

        for(int i = 1; i <= num1; i ++)
            sum1 += a[i];
        for(int i = num1 + 1; i <= N; i ++)
            sum2 += a[i];

        diff = sum2 - sum1;
    } else {
        num1 = num2 = N / 2;
        for(int i = 1; i <= num1; i ++)
            sum1 += a[i];
        for(int i = num1 + 1; i <= N; i ++)
            sum2 += a[i];

        diff = sum2 - sum1;
    }

    printf("Outgoing #: %d\n", num2);
    printf("Introverted #: %d\n", num1);
    printf("Diff = %lld\n", diff);
    return 0;
}