[题目] Luogu P3707 [SDOI2017]相关分析

参考资料:[Luogu 3707] SDOI2017 相关分析

P3707 [SDOI2017]相关分析

TFRAC FRAC DFRAC
\(\tfrac{\sum}{1}\) \(\frac{\sum}{1}\) \(\dfrac{\sum}{1}\)

\[\bar{x}=\frac{1}{R-L+1}\sum x_i​\]

\[\bar{y}=\frac{1}{R-L+1}\sum y_i​\]

\[\hat{a}=\dfrac{\sum_{i=L}^R(x_i-\bar{x})(y_i-\bar{y})}{\sum_{i=L}^R(x_i-\bar{x})^2}\]


\(\sum = \sum_{i=L}^{R}​\)

Query

\[Ans=\dfrac{ \sum x_iy_i-\bar{x}\sum y_i-\bar{y}\sum x_i+\sum \bar{x}\bar{y}}{\sum x_i^2-2\bar{x}\sum x_i+\sum\bar{x^2}}​\]

\[=\dfrac{ \sum x_iy_i-\frac{1}{R-L+1}\sum x_i\sum y_i-\frac{1}{R-L+1}\sum y_i\sum x_i+\sum \frac{1}{R-L+1}\sum x_i\frac{1}{R-L+1}\sum y_i}{\sum x_i^2-\frac{1}{R-L+1}2\sum x_i\sum x_i+\sum (\frac{1}{R-L+1}\sum x_i)^2}​\]

\[=\dfrac{ \sum x_iy_i-\frac{\sum x_i\sum y_i}{R-L+1}}{\sum x_i^2-\frac{(\sum x_i)^2}{R-L+1}}\]

下传Tag:先\(upd\)\(addX \quad addY\)

维护值:\(t1=\sum x​\) | \(t2=\sum y​\) | \(t3=\sum xy​\) | \(t4=\sum x^2​\)

维护Tag:\(addX \quad addY\)

\[\hat{a}=\dfrac{t3-\frac{t1t2}{R-L+1}}{t4-\frac{t1^2}{R-L+1}}​\]

Add

\[\Delta x = S \quad \Delta y= T​\]

\[\sum (x + S) = \sum x_i+(R-L+1)S ​\]

\[\sum(y+T)=\sum y_i+(R-L+1) T​\]

\[\sum(x+S)(y+T)=\sum(xy+Sy+Tx+ST)=\sum xy+S\sum y+T\sum x+(R-L+1)ST​\]

\[\sum(x+S)^2=\sum x^2+2S\sum x+(R-L+1)S^2\]

下传Tag:先\(upd\)\(addX \quad addY\)

顺序:先\(t3,t4\)\(t1,t2\)

Update

自然数平方和:\[\sum_{i=1}^ni=\frac{n(n+1)(2n+1)}{6}\]

1.\(\forall i \in [L,R]\quad x_i=i\ \ y_i=i​\)

\[\sum x=\sum y = \sum i = \frac{(R-L+1)(R+L)}{2}\]

\[\sum x^2 = \sum xy = \sum i ^2 = \sum_{i=1}^R i^2-\sum_{i=1}^{L-1}i^2=\frac{R(R+1)(2R+1)}{6}-\frac{L(L-1)(2L-1)}{6}\]

下传Tag:先\(upd\)\(addX \quad addY\)

清空Tag \(addX \quad addY\)

标记Tag \(upd​\)

2.ADD L R S T

Code

#include <iostream>
#include <cstdio>
#define ll long long

using namespace std;
const int N = 100005;
int n, m;
double X[N], Y[N];
struct Segment_Tree
{
    #define ls (p << 1)
    #define rs (p << 1 | 1)
    #define mid ((l + r) >> 1)
    bool upd[N << 2];
    double x[N << 2], y[N << 2];
    double t1[N << 2], t2[N << 2], t3[N << 2], t4[N << 2];
    inline void pushup(ll p)
    {
        t1[p] = t1[ls] + t1[rs];
        t2[p] = t2[ls] + t2[rs];
        t3[p] = t3[ls] + t3[rs];
        t4[p] = t4[ls] + t4[rs];
    }
    inline void pushdown(ll p, ll l, ll r)
    {
        double L = mid - l + 1, R = r - mid;
        if(upd[p]) {
            double Ll = l, Lr = mid, Rl = mid + 1, Rr = r;
            t1[ls] = t2[ls] = (Lr - Ll + 1.0) * (Lr + Ll) / 2.0;
            t1[rs] = t2[rs] = (Rr - Rl + 1.0) * (Rr + Rl) / 2.0;
            t3[ls] = t4[ls] = Lr * (Lr + 1.0) * (2.0 * Lr + 1.0) / 6.0 - Ll * (Ll - 1.0) * (2.0 * Ll - 1.0) / 6.0;
            t3[rs] = t4[rs] = Rr * (Rr + 1.0) * (2.0 * Rr + 1.0) / 6.0 - Rl * (Rl - 1.0) * (2.0 * Rl - 1.0) / 6.0;
            upd[ls] = upd[rs] = upd[p]; upd[p] = 0;
            x[ls] = x[rs] = y[ls] = y[rs] = 0;
        }
        if(x[p] || y[p]) {
            t3[ls] += x[p] * t2[ls] + y[p] * t1[ls] + L * x[p] * y[p];
            t3[rs] += x[p] * t2[rs] + y[p] * t1[rs] + R * x[p] * y[p];
        }
        if(x[p]) {
            t4[ls] += 2 * x[p] * t1[ls] + L * x[p] * x[p];
            t4[rs] += 2 * x[p] * t1[rs] + R * x[p] * x[p];
            t1[ls] += L * x[p];
            t1[rs] += R * x[p];
            x[ls] += x[p];
            x[rs] += x[p];
            x[p] = 0;
        }
        if(y[p]) {
            t2[ls] += (double)L * y[p];
            t2[rs] += (double)R * y[p];
            y[ls] += y[p];
            y[rs] += y[p];
            y[p] = 0;
        }
    }
    void build(ll p, ll l, ll r)
    {
        if(l == r) {
            t1[p] = X[l];
            t2[p] = Y[l];
            t3[p] = X[l] * Y[l];
            t4[p] = X[l] * X[l];
            return;
        }
        upd[p] = x[p] = y[p] = 0;
        build(ls, l, mid);
        build(rs, mid + 1, r);
        pushup(p);
    }
    void add(ll p, ll l, ll r, ll ql, ll qr, double S, double T)
    {
        if(ql <= l && r <= qr) {
            double len = (r - l + 1);
            t3[p] += S * t2[p] + T * t1[p] + len * S * T;
            t4[p] += 2 * S * t1[p] + len * S * S;
            t1[p] += len * S;
            t2[p] += len * T;
            x[p] += S; y[p] += T;
            return;
        }
        pushdown(p, l, r);
        if(ql <= mid) add(ls, l, mid, ql, qr, S, T);
        if(qr > mid) add(rs, mid + 1, r, ql, qr, S, T);
        pushup(p);
    }
    void update(ll p, ll l, ll r, ll ql, ll qr) {
        if(ql <= l && r <= qr) {
            t1[p] = t2[p] = (double)(r - l + 1.0) * (l + r) / 2.0;
            t3[p] = t4[p] = (double)r * (r + 1.0) * (2.0 * r + 1) / 6.0 - (double)l * (l - 1.0) * (2.0 * l - 1.0) / 6.0;
            x[p] = y[p] = 0;
            upd[p] = 1;
            return;
        }
        pushdown(p, l, r);
        if(ql <= mid) update(ls, l, mid, ql, qr);
        if(qr > mid) update(rs, mid + 1, r, ql, qr);
        pushup(p);
    }
    double query(ll p, ll l, ll r, ll ql, ll qr, ll f) {
        if(ql <= l && r <= qr) {
            if(f == 1) return t1[p];
            if(f == 2) return t2[p];
            if(f == 3) return t3[p];
            if(f == 4) return t4[p];
        }
        pushdown(p, l, r);
        double res = 0;
        if(ql <= mid) res += query(ls, l, mid, ql, qr, f);
        if(qr > mid) res += query(rs, mid + 1, r, ql ,qr, f);
        return res;
    }
    #undef ls
    #undef rs
    #undef mid
};
struct Segment_Tree Tree;
int main()
{
    scanf("%d %d", &n, &m);
    for(int i = 1; i <= n; i++) scanf("%lf", &X[i]);
    for(int i = 1; i <= n; i++) scanf("%lf", &Y[i]);
    Tree.build(1, 1, n);
    int opt, L, R; double S, T;
    while(m--)
    {
        scanf("%d", &opt);
        if(opt == 1)
        {
            scanf("%d %d", &L, &R);
            double t1 = Tree.query(1, 1, n, L, R, 1);
            double t2 = Tree.query(1, 1, n, L, R, 2);
            double t3 = Tree.query(1, 1, n, L, R, 3);
            double t4 = Tree.query(1, 1, n, L, R, 4);
            double a_ = (t3 - (t1 * t2) / (double)(R - L + 1)) / (t4 - (t1 * t1) / (double)(R - L + 1));
            printf("%.10lf\n", a_);
        }
        else if(opt == 2)
        {
            scanf("%d %d %lf %lf", &L, &R, &S, &T);
            Tree.add(1, 1, n, L, R, S, T);
        }
        else if(opt == 3)
        {
            scanf("%d %d %lf %lf", &L, &R, &S, &T);
            Tree.update(1, 1, n, L, R);
            Tree.add(1, 1, n, L, R, S, T);
        }
    }
    return 0;
}

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转载自www.cnblogs.com/colorfulmist/p/10623792.html