L2-017 人以群分 (25分)[PTA][奇怪的模拟题]

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

 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

/2再排一个序

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<set>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
const int maxn = 110000;

ll ac[maxn];

int main()
{
    int N;
    cin>>N;
    for(int i=0;i<N;i++)
    {
        cin>>ac[i];
    }
    sort(ac,ac+N);
    int A=N/2;
    int B=N-A;
    int flag=0;
    ll max=0,min=0;
    for(int i=N-1;flag<B;i--)
    {
        flag++;
        max+=ac[i];
    }
    flag=0;
    for(int i=0;flag<A;i++)
    {
        flag++;
        min+=ac[i];
    }
    cout<<"Outgoing #: "<<B<<endl;
    cout<<"Introverted #: "<<A<<endl;
    cout<<"Diff = "<<max-min;
    return 0;
}


 参考:https://blog.csdn.net/cuideman/article/details/65956128?ops_request_misc=%257B%2522request%255Fid%2522%253A%2522160635406319215668835802%2522%252C%2522scm%2522%253A%252220140713.130102334.pc%255Fblog.%2522%257D&request_id=160635406319215668835802&biz_id=0&utm_medium=distribute.pc_search_result.none-task-blog-2~blog~first_rank_v1~rank_blog_v1-25-65956128.pc_v1_rank_blog_v1&utm_term=L2-017+人以群分++25分&spm=1018.2118.3001.4450

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转载自blog.csdn.net/qq_43660826/article/details/110162115