【高手训练】【树状数组】电子速度

<高手训练例题>

正片开始

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：

\[\sum_{1\leq i \leq j \leq n} |v_i \times v_j|^2(\times表示叉乘) \]

电子的速度常常会发生变化。也就是说，有两种类型的操作：

• \(1\ \ p\ \ x\ \ y\)将 \(p\) 改为\((x,\ y)\)；

• \(2\ \ l\ \ r\)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：

\[\begin{aligned} ans\ &= \sum_{1 \leq i < j \leq n} \mathrm{(x_iy_j - x_jy_i)^2} \\ &=\sum_{1 \leq i < j \leq n} \mathrm{x_i^2y_j^2} + \sum_{1 \leq i < j \leq n} \mathrm{x_j^2y_i^2} - 2\sum_{1 \leq i < j \leq n}\mathrm{x_ix_jy_iy_j}\\ &=\sum_{1 \leq i, j \leq n}\mathrm{[i ≠j]x_i^2y_j^2} - \sum_{1 \leq i,j \leq n}\mathrm{[i ≠ j] (x_iy_i \cdot x_jy_j)} \\ &= (\sum_{i = 1}^{n}\mathrm{x_i^2} \cdot \sum_{i = 1}^{n}\mathrm{y_i^2} - \sum_{i = 1}^{n}\mathrm{(x_i^2y_i^2) })- \mathrm{(\sum_{i = 1}^{n} x_iy_i \cdot \sum_{i = 1}^{n}x_iy_i - \sum_{i = 1}^{n}(x_iy_i)^2)} \\ &=\mathrm{\sum_{i = 1}^{n}x_i^2 \cdot \sum_{i = 1}^{n}y_i^2 - (\sum_{i = 1}^{n} x_iy_i)^2} \end{aligned} \]

三个树状数组维护三个值\(\sum_{i = 1}^n x_i^2\) 、 \(\sum_{i = 1}^n y_i^2\) 、 \(\sum_{i = 1}^nx_iy_j\)，三个值。这样就可以支持修改操作并快速维护答案了。

\(\mathrm{Code:}\)

#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;
}