标签：reset lse sort its ref ace problems list can
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??。
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。
10
23 8 10 99 46 2333 46 1 666 555
Outgoing #: 5
Introverted #: 5
Diff = 3611
13
110 79 218 69 3721 100 29 135 2 6 13 5188 85
Outgoing #: 7 Introverted #: 6 Diff = 9359
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;
}
标签：reset lse sort its ref ace problems list can
原文地址：https://www.cnblogs.com/zlrrrr/p/9649073.html
