【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_{1leq i leq j leq n} |v_i imes v_j|^2( imes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：

[egin{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} - 2sum_{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;
}