天才绅士少女助手克里斯蒂娜 [数学+树状数组]

$天才绅士少女助手克里斯蒂娜$

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$\color{blue}{最初想法}$

$\begin{aligned}&\ \ \ \ \sum_{i=l}^{r}\sum_{j=i+1}^r(x_iy_j-x_jy_i)^2 \\ &= \sum_{i=l}^rx_i^2\sum_{j=i+1}^ry_j^2 + \sum_{i=l}^ry_i^2\sum_{j=i+1}^rx_j^2-2\sum_{i=l}^rx_iy_i\sum_{j=i+1}^rx_jy_j\\ \end{aligned}$

然后处理前缀和, 时间复杂度 $O(N^2)$ .

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$\color{red}{正解部分}$

接着上式化简得

$\begin{aligned} &\ \ \ \ \sum_{i=l}^rx_i^2\sum_{j=i+1}^ry_j^2 + \sum_{i=l}^ry_i^2\sum_{j=i+1}^rx_j^2-2\sum_{i=l}^rx_iy_i\sum_{j=i+1}^rx_jy_j\\ &= \sum_{i=l}^rx_i^2\sum_{j=i+1}^ry_j^2 + \sum_{i=l}^rx_i^2\sum_{j=l}^{i-1}y_j^2 - (\sum_{i=l}^rx_iy_i\sum_{j=l}^rx_jy_j - \sum_{i=l}^r x_i^2y_i^2)\\ &= \sum_{i=l}^rx_i^2(\sum_{j=l}^ry_j^2 - y_i^2) - \sum_{i=l}^rx_iy_i\sum_{j=l}^rx_jy_j + \sum_{i=l}^r x_i^2y_i^2\\ &= \sum_{i=l}^rx_i^2\sum_{j=l}^ry_j^2 - \sum_{i=l}^rx_i^2y_i^2 - \sum_{i=l}^rx_iy_i\sum_{j=l}^rx_jy_j + \sum_{i=l}^r x_i^2y_i^2 \\ &= \sum_{i=l}^rx_i^2\sum_{j=l}^ry_j^2 - \sum_{i=l}^rx_iy_i\sum_{j=l}^rx_jy_j \end{aligned}$

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$\color{red}{实现部分}$

使用树状数组维护即可 .

#include<bits/stdc++.h>
#define reg register
typedef long long ll;

int read(){
        char c;
        int s = 0, flag = 1;
        while((c=getchar()) && !isdigit(c))
                if(c == '-'){ flag = -1, c = getchar(); break ; }
        while(isdigit(c)) s = s*10 + c-'0', c = getchar();
        return s * flag;
}

const int maxn = 1000005;
const int mod = 20170927;

int N;
int M;

ll v[maxn][2];

struct Bit_Tree{
        ll v[maxn];
        void Add(int k, ll p){ p%=mod; while(k<=N)v[k]+=p,k+=k&-k; }
        ll Query(int k){ ll s = 0; while(k)s+=v[k],s%=mod,k-=k&-k; return s; }
} bit_t[3];

void Work(){
        int opt = read();
        if(opt == 1){
                int pos = read();
                bit_t[0].Add(pos, -v[pos][0]*v[pos][0]);
                bit_t[1].Add(pos, -v[pos][1]*v[pos][1]);
                bit_t[2].Add(pos, -v[pos][0]*v[pos][1]);
                v[pos][0] = read(), v[pos][1] = read();
                bit_t[0].Add(pos, v[pos][0]*v[pos][0]);
                bit_t[1].Add(pos, v[pos][1]*v[pos][1]);
                bit_t[2].Add(pos, v[pos][0]*v[pos][1]);
        }else{
                int l = read(), r = read();
                ll Tmp_1 = bit_t[0].Query(r) - bit_t[0].Query(l-1);
                ll Tmp_2 = bit_t[1].Query(r) - bit_t[1].Query(l-1);
                ll Tmp_3 = bit_t[2].Query(r) - bit_t[2].Query(l-1);
                Tmp_1 = (Tmp_1 % mod + mod) % mod;
                Tmp_2 = (Tmp_2 % mod + mod) % mod;
                Tmp_3 = (Tmp_3 % mod + mod) % mod;
                printf("%lld\n", ( (Tmp_1*Tmp_2%mod) - (Tmp_3*Tmp_3%mod) + mod ) % mod);
        }
}

int main(){
        N = read(), M = read();
        for(reg int i = 1; i <= N; i ++) v[i][0] = read(), v[i][1] = read(); 
        for(reg int i = 1; i <= N; i ++){
                bit_t[0].Add(i, v[i][0]*v[i][0]);
                bit_t[1].Add(i, v[i][1]*v[i][1]);
                bit_t[2].Add(i, v[i][0]*v[i][1]);
        }
        while(M --) Work();
        return 0;
}