【Codeforces837G】Functions On The Segments

这题只要读懂了题意，开大了范围，看清了如何强制在线，就能很快AC（雾）

#include <bits/stdc++.h>#define gc getchar()#define mid (l+r>>1)#define ll long long#define mod 1000000000#define M 30000009#define N 200009using namespace std;ll n,tr[M][2],root[N],cnt,lson[M],rson[M];ll read(){    ll x=1;    char ch;    while (ch=gc,ch<'0'||ch>'9') if (ch=='-') x=-1;    ll s=ch-'0';    while (ch=gc,ch>='0'&&ch<='9') s=s*10+ch-'0';    return s*x;}void ins(ll &cur,ll l,ll r,ll x,ll y,ll k,ll last){    cur=++cnt;    lson[cur]=lson[last],rson[cur]=rson[last];    tr[cur][k]=tr[last][k]+y;    tr[cur][k^1]=tr[last][k^1];    if (l==r) return;    if (x<=mid) ins(lson[cur],l,mid,x,y,k,lson[last]);    else ins(rson[cur],mid+1,r,x,y,k,rson[last]);}ll qry(ll cur,ll l,ll r,ll L,ll R,ll k){    if (!cur) return 0;    if (L<=l&&R>=r) return tr[cur][k];    ll ret=0;    if (L<=mid) ret+=qry(lson[cur],l,mid,L,R,k);    if (R>mid) ret+=qry(rson[cur],mid+1,r,L,R,k);    return ret;}int main(){    n=read();    for (ll i=1;i<=n;i++)    {        ll x1=read(),x2=read(),y1=read(),a=read(),b=read(),y2=read();        ins(root[i],0,mod,0,y1,0,root[i-1]);        ins(root[i],0,mod,x1+1,b-y1,0,root[i]);        ins(root[i],0,mod,x2+1,y2-b,0,root[i]);        ins(root[i],0,mod,x1+1,a,1,root[i]);        ins(root[i],0,mod,x2+1,-a,1,root[i]);    }    ll Ans=0;    ll q=read();    for (int i=1;i<=q;i++)    {        ll l=read(),r=read(),x=(read()+Ans)%mod;        ll Ans1=qry(root[r],0,mod,0,x,0)-qry(root[l-1],0,mod,0,x,0);        ll Ans2=qry(root[r],0,mod,0,x,1)-qry(root[l-1],0,mod,0,x,1);        //if (i==4999) x=200001;        printf("%lld\n",Ans=(Ans1+x*Ans2));    }    return 0;}