hdu 5073

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5073

思路：一开始忘了排序，wa了好几发。。。选择区间长度为N - K的连续的数， 然后其余的K个数都移动到这N-K个数的中心就可以，公式为ans = min(ans, pos[i] - center_point) ^ 2），把这个公式展开然后预处理一下就可以了，复杂度为O(N)。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

const int MAX_N = (50000 + 5000);
int N, K;
double pos[MAX_N], sum1[MAX_N], sum2[MAX_N];

int main()
{
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        scanf("%d %d", &N, &K);

        for (int i = 1; i <= N; ++i) {
            scanf("%lf", &pos[i]);
        }

        if (N == K) {
            puts("0");
            continue;
        }

        sort(pos + 1, pos + 1 + N);

        sum1[0] = sum2[0] = 0;
        for (int i = 1; i <= N; ++i) {
            sum1[i] = sum1[i - 1] + pos[i];
            sum2[i] = sum2[i - 1] + pos[i] * pos[i];
        }


        double ans = 1e100, point_b = 0;
        int M = N - K;

        for (int i = M; i <= N; ++i) {

            point_b = (sum1[i] - sum1[i - M]) / M;

            double tmp = (sum2[i] - sum2[i - M])  - 2 * point_b * (sum1[i] - sum1[i - M]) + M * point_b * point_b;

            ans = min(ans, tmp);
        }

        printf("%.12f
", ans);
    }
    return 0;
}