【高手训练】【树状数组】电子速度

题目

选取显像管的任意一个平面，一开始平面内有$n$个电子，初始速度分别为$v_i$，定义飘升系数为：
$$\sum _{1\le i\le j\le n}|v_i\times v_j{|}^2(\times 表示叉乘)$$
电子的速度常常会发生变化。也就是说，有两种类型的操作：

• $1pxy$将 $p$ 改为$(x,y)$；

• $2lr$询问 $[l,r]$ 这段区间内的电子的飘升系数。

答案对 $20170927$ 取模即可。

Solution

观察答案并推导：
$$\begin{array}{rl}ans& =\sum _{1\le i<j\le n}({\mathrm{x}}_{\mathrm{i}}{\mathrm{y}}_{\mathrm{j}}-{\mathrm{x}}_{\mathrm{j}}{\mathrm{y}}_{\mathrm{i}}{)}^2\\ & =\sum _{1\le i<j\le n}{\mathrm{x}}_{\mathrm{i}}^2{\mathrm{y}}_{\mathrm{j}}^2+\sum _{1\le i<j\le n}{\mathrm{x}}_{\mathrm{j}}^2{\mathrm{y}}_{\mathrm{i}}^2-2\sum _{1\le i<j\le n}{\mathrm{x}}_{\mathrm{i}}{\mathrm{x}}_{\mathrm{j}}{\mathrm{y}}_{\mathrm{i}}{\mathrm{y}}_{\mathrm{j}}\\ & =\sum _{1\le i,j\le n}[\mathrm{i}\ne \mathrm{j}]{\mathrm{x}}_{\mathrm{i}}^2{\mathrm{y}}_{\mathrm{j}}^2-\sum _{1\le i,j\le n}[\mathrm{i}\ne \mathrm{j}]({\mathrm{x}}_{\mathrm{i}}{\mathrm{y}}_{\mathrm{i}}\cdot {\mathrm{x}}_{\mathrm{j}}{\mathrm{y}}_{\mathrm{j}})\\ & =(\sum _{i=1}^n{x}_i^2\cdot \sum _{i=1}^n{y}_i^2-\sum _{i=1}^n(x_i^2{y}_i^2))-(\sum _{i=1}^n{x}_i{y}_i\cdot \sum _{i=1}^n{x}_i{y}_i-\sum _{i=1}^n(x_i{y}_i{)}^2)\\ & =\sum _{i=1}^n{x}_i^2\cdot \sum _{i=1}^n{y}_i^2-(\sum _{i=1}^n{x}_i{y}_i{)}^2\end{array}$$
三个树状数组维护三个值${\sum }_{i=1}^n{x}_i^2$ 、 ${\sum }_{i=1}^n{y}_i^2$ 、 ${\sum }_{i=1}^n{x}_i{y}_j$，三个值。这样就可以支持修改操作并快速维护答案了。

$\mathrm{C}\mathrm{o}\mathrm{d}\mathrm{e}:$

#include <bits/stdc++.h>
#define N 1000110
#define mod 20170927
using namespace std;
int n, m;

inline int add(int a, int b) {
    if (b < 0)
        return (a + b < 0 ? a + b + mod : a + b);
    else
        return (a + b >= mod ? a + b - mod : a + b);
}// add函数把加减写一起
inline int del(int a, int b) { return a - b < 0 ? a - b + mod : a - b; }
inline int mul(int a, int b) { return 1LL * a * b % mod; }

inline int read() {
    int s = 0, w = 1;
    char c = getchar();
    while ((c > '9' || c < '0') && c != '-') c = getchar();
    if (c == '-') w = -1, c = getchar();
    while (c <= '9' && c >= '0')
        s = (s << 3) + (s << 1) + c - '0', c = getchar();
    return s * w;
}
void write(int x) {
    if (x < 0) x = ~x + 1, putchar('-');
    if (x > 9) write(x / 10);
    putchar(x % 10 + 48);
    return void();
}

struct vec {
    int x, y;
    vec() {}
    vec(int _x, int _y) {
        x = _x;
        y = _y;
    }
    inline vec operator-() { return vec(-x, -y); }
} v[N];
struct BIT {
    int a[N], b[N], c[N];
    void Inc(int x, vec v, int w) { //w为加&减的处理。
        for (; x <= n; x += x & (-x)) {
            a[x] = add(a[x], w * mul(v.x, v.x));
            b[x] = add(b[x], w * mul(v.y, v.y));
            c[x] = add(c[x], w * mul(v.x, v.y));
        }
    }
    int aska(int x) {
        int sum = 0;
        for (; x; x -= x & (-x)) sum = add(sum, a[x]);
        return sum;
    }
    int askb(int x) {
        int sum = 0;
        for (; x; x -= x & (-x)) sum = add(sum, b[x]);
        return sum;
    }
    int askc(int x) {
        int sum = 0;
        for (; x; x -= x & (-x)) sum = add(sum, c[x]);
        return sum;
    }
    int Ask(int l, int r) {
        int s1 = del(aska(r), aska(l - 1));
        int s2 = del(askb(r), askb(l - 1));
        int s3 = del(askc(r), askc(l - 1));
        return del(mul(s1, s2), mul(s3, s3));
    }
} B;
signed main() {
    n = read();
    m = read();
    for (int i = 1; i <= n; ++i) {
        int x = read(), y = read();
        v[i] = vec(x, y);
        B.Inc(i, v[i], 1);
    }
    for (int i = 1; i <= m; ++i) {
        int opt = read();
        if (opt == 1) {
            int p = read(), x = read(), y = read();
            vec tmp = vec(x, y);
            B.Inc(p, v[p], -1);
            B.Inc(p, tmp, 1);
            v[p] = tmp;
        }
        if (opt == 2) {
            int x = read(), y = read();
            write(B.Ask(x, y));
            putchar(10);
        }
    }
    return 0;
}