标签：lan html its sizeof rip printf 正整数 sort diff
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≤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??。
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
机智的我双开数组代表 + if 讨论 1A
鲁老师的学生不许抄我的，我会被挂的。
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL acv[100100],sum[100100];
int main()
{
memset(acv,0,sizeof(acv));
memset(sum,0,sizeof(sum));
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%lld",&acv[i]);
}
sort(acv+1,acv+1+n);
sum[0]=0;
for(int i=1;i<=n;i++){
sum[i] = sum[i-1]+acv[i];
}
int a1 = n/2;
int b1 = n-n/2;
LL diff1 = 0;
diff1 = sum[n] - 2*sum[a1];
int b2 = n/2;
int a2 = n-n/2;
LL diff2 = 0;
diff2 = sum[n] - 2*sum[a2];
if(diff1>diff2){
printf("Outgoing #: %d\n",b1);
printf("Introverted #: %d\n",a1);
printf("Diff = %lld\n",diff1);
}
else
{
printf("Outgoing #: %d\n",b2);
printf("Introverted #: %d\n",a2);
printf("Diff = %lld\n",diff2);
}
}
标签：lan html its sizeof rip printf 正整数 sort diff
原文地址：http://www.cnblogs.com/masterchd/p/7886615.html
