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:
13
110 79 218 69 3721 100 29 135 2 6 13 5188 85
立个flag,下学期好好做人。。。我爱学习,学习使我快乐?
最简单的分类情况,分奇偶,注意个数即可,如果是奇数,把中间的那个数放到外向型里边,这样才能保证最后的diff的值最大
#include <iostream>
#include <algorithm>
using namespace std;
int main(){
ios::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
int n;
cin >> n;
int a[n];
for(int i = 0;i<n;i++){
cin >> a[i];
}
sort(a,a+n);
if(n%2==0){
int out[n/2],in[n/2];
int sum1 = 0,sum2 = 0;
int k = 0;
for(int i = 0;i<n/2;i++){
in[k++] = a[i];
sum1 += a[i];
}
k = 0;
for(int i = n/2;i<n;i++){
out[k++] = a[i];
sum2 += a[i];
}
cout << "Outgoing #: " << n/2 << endl;
cout << "Introverted #: " << n/2 << endl;
cout << "Diff = " << sum2-sum1 << endl;
}else{
int out[n/2+1],in[n/2];
int sum1 = 0,sum2 = 0;
int k = 0;
for(int i = 0;i<n/2;i++){
in[k++] = a[i];
sum1 += a[i];
}
k = 0;
for(int i = n/2;i<n;i++){
out[k++] = a[i];
sum2 += a[i];
}
cout << "Outgoing #: " << n/2+1 << endl;
cout << "Introverted #: " << n/2 << endl;
cout << "Diff = " << sum2-sum1 << endl;
}
return 0;
}