$Solution:$

一开始没看懂题。后来大致理解了一下，所谓的v是一个二维向量，有x和y两个参数。那个$ imes$是叉乘，即$(x_i y_j-x_j y_i)$。

所以题意就是给你一个x序列和y序列，对于每次询问的区间$[l,r]$，求$sum limits _{lleq i<j leq r}(x_iy_j-x_jy_i)^2$。带修。

点对的形式比较麻烦，尝试化成单点的柿子：

先拆平方：

$sum limits _{lleq i<j leq r}x_i^2y_j^2-2x_ix_jy_iy_j+x_j^2y_i^2$

大力化简：

$large egin{array}{ll} ans &=& sum limits_{i=l}^{r} sum limits_{j=i+1}^r (x_i^2y_j^2+x_j^2y_i^2-2x_iy_ix_jy_j) \ &=& sum limits_{i=l}^r sum limits_{j=i+1}^r x_i^2y_j^2 + sum limits_{i=l}^r sum limits_{j=i+1}^r x_j^2y_i^2 - sum limits_{i=l}^r sum limits_{j=i+1}^r 2x_iy_ix_jy_j \ &=& sum limits_{i=l}^r sum limits_{j=l}^r [i eq j]*x_i^2y_j^2 - sum limits_{i=l}^r sum limits_{j=l}^r [i eq j]*x_iy_ix_jy_j \ &=& sum limits_{i=l}^r x_i^2 (sum limits_{j=l}^r y_j^2 -y_i^2) - (sum limits_{i=l}^r x_iy_i (sum limits_{j=l}^r x_jy_j - x_iy_i)) \ &=& sum limits_{i=l}^r x_i^2 sum limits_{j=l}^r y_j^2 - sum limits_{i=l}^r x_i^2y_i^2 - (sum limits_{i=l}^r x_iy_i sum limits_{j=l}^r x_jy_j - sum limits_{i=l}^r x_i^2 y_i^2) \ &=& sum limits_{i=l}^r x_i^2* sum limits_{i=l}^ry_i^2 - (sum limits_{i=l}^r x_iy_i)^2end{array}$

然后开三个树状数组分别维护${x_i}^2,{y_i}^2,x_i y_i$即可。

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
#define pa pair<ll,ll>
typedef long long ll;
const ll mod=20170927;
ll read()
{
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return x*f;
}
const int N=1e6+5;
int n,m;
ll c[3][N];
pa v[N];
int lb(int x){return x&-x;}
void add(int x,ll val,int id)
{
    for( ;x<=n;x+=lb(x))
        c[id][x]+=val,(c[id][x]+=mod)%=mod;
}
ll query(int x,int id)
{
    ll res=0;
    for( ;x;x-=lb(x))
        (res+=c[id][x])%=mod;
    return (res+mod)%mod;
}
ll ask(int l,int r,int id)
{
    return (query(r,id)-query(l-1,id)+mod)%mod;
}
int main()
{
    n=read();m=read();
    for(int i=1;i<=n;i++)
    {
        v[i].first=read(),v[i].second=read();
        add(i,v[i].first*v[i].first,0);
        add(i,v[i].second*v[i].second,1);
        add(i,v[i].first*v[i].second,2);
    }

    while(m--)
    {
        int op=read();
        if(op==1)
        {
            int pos=read();
            ll x=read(),y=read();
            add(pos,x*x-v[pos].first*v[pos].first,0);
            add(pos,y*y-v[pos].second*v[pos].second,1);
            add(pos,x*y-v[pos].first*v[pos].second,2);
            v[pos].first=x;v[pos].second=y;

        }
        if(op==2)
        {
            int l=read(),r=read();
            ll res=ask(l,r,0)*ask(l,r,1)%mod,res1=ask(l,r,2);
            res=(res-(res1*res1%mod)+mod)%mod;
            printf("%lld
",res);
        }
    }
    return 0;
}

[信息学奥赛一本通oj1741]电子速度 题解

$Solution:$

[信息学奥赛一本通oj1741]电子速度题解