题目链接：
https://acm.ecnu.edu.cn/contest/140/problem/F/
题目：
思路：
因为方差是用来评估数据的离散程度的，因此最优的m个数一定是排序后连续的，所以我们可以先排序然后对每m个连续的数取个min。
代码实现如下：
#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std; typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL; #define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0) const double eps = 1e-;
const int mod = ;
const int maxn = 1e6 + ;
const double pi = acos(-);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL; int n, m;
int a[maxn];
LL sum1[maxn], sum2[maxn]; int main(){
scanf("%d%d", &n, &m);
for(int i = ; i <= n; i++) {
scanf("%d", &a[i]);
}
sort(a + , a + n + );
for(int i = ; i <= n; i++) {
sum1[i] = sum1[i-] + a[i];
sum2[i] = sum2[i-] + a[i] * a[i];
}
LL ans = INF;
for(int i = ; i <= n - m + ; i++) {
ans = min(ans, m * (sum2[i+m-] - sum2[i-]) - (sum1[i+m-] - sum1[i-]) * (sum1[i+m-] - sum1[i-]));
}
printf("%lld\n", ans);
return ;
}