「Luogu P6069」[MdOI2020] Group

Portal

Portal1: Luogu

Solution

我们先将题目的式子简化一下：

\[\begin{aligned}S&=\frac{1}{n} \sum_{i=1}^n(a_i-p)^2\\S&=\frac{1}{n} \sum_{i=1}^n(a_i^2-2a_ip+p^2)\\S&=\frac{1}{n} \sum_{i=1}^n\left(a_i^2-2a_i\times\frac{sum1}{n}+\left(\frac{sum1}{n}\right)^2\right)\\nS&=sum2+\frac{sum1^2}{n}-2\times\frac{sum1^2}{n}\\nS&=sum2-\frac{sum1^2}{n}\end{aligned} \]

不难发现，要先排序，然后选取一段连续区间的方差一定比选取同样个数的不连续区间的方差小。所以我们可以二分答案（区间的长度），然后连续取每一段区间逐一判断就好了，时间复杂度为\(\mathbb{O(n \log n)}\)。

修改的数不用管，因为你可以修改成任意一个数。

Code

#include<bits/stdc++.h>
#define int __int128
//不开int128会炸
using namespace std;

typedef long long LL;
const int MAXN = 200005;
int n, m, a[MAXN], sum1[MAXN], sum2[MAXN];
LL ans;
inline int read() {//int128必须快读
    char ch = getchar();
    int x = 0, f = 1;
    while (ch < '0' || ch > '9') {
        if (ch == '-') f = -1;
        ch = getchar();
    }
    while ('0' <= ch && ch <= '9') {
        x = (x << 1) + (x << 3) + ch - '0';
        ch = getchar();
    }
    return x * f;
}
inline void prepare() {
    sort(a + 1, a + n + 1);
    for (LL i = 1; i <= n; i++) {
        sum1[i] = sum1[i - 1] + a[i];
        sum2[i] = sum2[i - 1] + a[i] * a[i];
    }
}
inline bool check(LL x) {
    for (LL L = 1, R = L + x - 1; R <= n; L++, R++)
        if ((sum2[R] - sum2[L - 1]) * x - (sum1[R] - sum1[L - 1]) * (sum1[R] - sum1[L - 1]) <= x * m) return 1;
    return 0;
}
signed main() {
    n = read(), m = read();
    for (LL i = 1; i <= n; i++)
        a[i] = read();
    prepare();
    LL l = 1, r = n;
    while (l <= r) {
        LL mid = l + r >> 1;
        if (check(mid)) l = mid + 1, ans = n - mid; else r = mid - 1;
    }
    printf("%lld\n", ans);
    return 0;
}