牛客寒假训练营1__小a与204

原题链接:https://ac.nowcoder.com/acm/contest/317/B

来源:牛客网

题目描述:

法1:利用贪心的思想,从小到大排序,先选大的,也就是4,再选小的,也就是0

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cmath>
using namespace std;
typedef long long ll;
const int N=1e5+10;
int a[N], n;
int main()
{
    ll ans = 0;
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i)
        scanf("%d", &a[i]);
    a[0] = 0;
    sort(a, a+n+1);
    for(int i = n; i > n/2; --i)
        ans += pow(a[i]-a[n-i], 2)+pow(a[n-i+1]-a[i], 2);
    printf("%lld\n", ans);
    return 0;
}

法二,某大佬写的浅显易懂的代码:

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cmath>
using namespace std;
typedef long long ll;
const int N=1e5+10;
int a[N];
int main()
{
    int n;
    ll ans = 0;
    cin >> n;
    for(int i = 1; i <= n; ++i)
        cin >> a[i];
    a[0] = 0;
    sort(a+1, a+n+1);
    int l = 1, r = n;
    while(l<r)
    {
        ans += pow(a[r]-a[l], 2);
        r--;
        ans += pow(a[r]-a[l], 2);
        l++;
    }
    ans += a[n]*a[n];
    printf("%lld\n", ans);
    return 0;
}

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转载自blog.csdn.net/qq_31038073/article/details/86664880