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\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P1471 方差

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P1471 方差

开始没有看到输入可以是实数，就全部 $RE$ 了。

方差：
$$
\begin {aligned}
s ^ 2 &= \frac 1 n (\sum _ {i = 1} ^ n (A _ i - \bar A) ^ 2)\\
&= \frac 1 n [\sum _ {i = 1} ^ n (A_ i ^ 2 - 2A_i\bar A + \bar A ^ 2)]\\
&= \frac 1 n (\sum _ {i = 1} ^ n A_ i ^ 2 - 2\bar A\sum _ {i = 1} ^ n A_ i + \sum _ {i = 1} ^ n\bar A ^ 2)\\
&= \frac 1 n (\sum _ {i = 1} ^ n A_ i ^ 2 - 2\frac {(\sum _ {i = 1} ^ n A_ i) ^ 2} n + \frac {(\sum _ {i = 1} ^ n A_ i) ^ 2} n)\\
&= \frac 1 n \sum _ {i = 1} ^ n A_ i ^ 2 - \bar A ^ 2
\end {aligned}
$$
所以线段树里维护和、平方和即可。

区间修改维护平方和：
$$
\begin {aligned}
\sum _ {i = 1} ^ n (A _ i + k) ^ 2 & = \sum _ {i = 1} ^ n (A _ i ^ 2 - 2kA_i + k ^ 2) \\
& = \sum _ {i = 1} ^ n A _i ^2 - 2k \sum _ {i = 1} ^ n A _ i + nk ^ 2
\end {aligned}
$$

查看代码 
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#include <cstdio>
using namespace std;
const int N = 1e5 + 10;
int n, m;
double w[N];
struct Node
{
    int l, r;
    double s, s2, tag;
} tr[N << 2];
void cal(Node &x, double k)
{
    x.s2 += 2 * k * x.s + (x.r - x.l + 1) * k * k;
    x.s += (x.r - x.l + 1) * k;
    x.tag += k;
}
void pushup(Node &x, Node l, Node r)
{
    x.l = l.l, x.r = r.r;
    x.s = l.s + r.s;
    x.s2 = l.s2 + r.s2;
}
void pushdown(int x)
{
    cal(tr[x << 1], tr[x].tag);
    cal(tr[x << 1 | 1], tr[x].tag);
    tr[x].tag = 0;
}
void build(int x, int l, int r)
{
    if (l == r)
    {
        tr[x].l = tr[x].r = l;
        tr[x].s = w[l];
        tr[x].s2 = w[l] * w[l];
        return;
    }
    int mid = l + r >> 1;
    build(x << 1, l, mid);
    build(x << 1 | 1, mid + 1, r);
    pushup(tr[x], tr[x << 1], tr[x << 1 | 1]);
}
void modify(int x, int l, int r, double k)
{
    if (tr[x].l >= l && tr[x].r <= r)
    {
        cal(tr[x], k);
        return;
    }
    if (tr[x].tag)
        pushdown(x);
    int mid = tr[x].l + tr[x].r >> 1;
    if (l <= mid)
        modify(x << 1, l, r, k);
    if (r > mid)
        modify(x << 1 | 1, l, r, k);
    pushup(tr[x], tr[x << 1], tr[x << 1 | 1]);
}
Node query(int x, int l, int r)
{
    if (tr[x].l >= l && tr[x].r <= r)
        return tr[x];
    if (tr[x].tag)
        pushdown(x);
    int mid = tr[x].l + tr[x].r >> 1;
    if (r <= mid)
        return query(x << 1, l, r);
    if (l > mid)
        return query(x << 1 | 1, l, r);
    Node res;
    pushup(res, query(x << 1, l, r), query(x << 1 | 1, l, r));
    return res;
}
int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
        scanf("%lf", &w[i]);
    build(1, 1, n);
    for (int op, l, r; m; m--)
    {
        scanf("%d%d%d", &op, &l, &r);
        if (op == 1)
        {
            double k;
            scanf("%lf", &k);
            modify(1, l, r, k);
            continue;
        }
        Node t = query(1, l, r);
        double avrg = (double)t.s / (t.r - t.l + 1);
        if (op == 2)
            printf("%.4lf\n", avrg);
        else
            printf("%.4lf\n", (double)t.s2 / (t.r - t.l + 1) - avrg * avrg);
    }
    return 0;
}
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