BZOJ2874 : 训练士兵
设$a[i][j]$表示$(i,j)$右下角要增加多少
$aj[i][j]=a[i][j]\times j$
$ai[i][j]=a[i][j]\times i$
$aij[i][j]=a[i][j]\times i\times j$
则查询$(x,y)$左上角内的权值和时,
答案$=(x+1)(y+1)ask_{a}(x,y)-(x+1)ask_{aj}(x,y)-(y+1)ask_{ai}(x,y)+ask_{aij}(x,y)$。
首先将坐标离散化,将修改拆成4个单点修改,然后从左往右插入点,用可持久化线段树维护二维前缀和。
查询时拆成4次二维前缀和查询即可。
时间复杂度$O((k+q)\log k)$。
#include<cstdio> #include<algorithm> using namespace std; typedef long long ll; const int N=80010,M=3000000; int n,m,k,q,i,j,x,y,X1,X2,Y1,Y2,s,cnt,bx[N],by[N],cl,g[N],nxt[N<<1];ll ans; struct P{int x,y,z;P(){}P(int _x,int _y,int _z){x=_x,y=_y,z=_z;}}a[N<<1]; int tot,T[N],l[M],r[M]; struct V{ ll o,j,i,ij; V(){o=j=i=ij=0;} V(ll _o,ll _j,ll _i,ll _ij){o=_o,j=_j,i=_i,ij=_ij;} V(int x,int y,int z){o=z,j=1LL*z*y,i=1LL*z*x,ij=1LL*z*x*y;} inline V operator+(const V&b){return V(o+b.o,j+b.j,i+b.i,ij+b.ij);} }v[M],p; inline int lowerx(int x){ int l=1,r=cl,mid,t=0; while(l<=r)if(bx[mid=(l+r)>>1]<=x)l=(t=mid)+1;else r=mid-1; return t; } inline int lowery(int x){ int l=1,r=cl,mid,t=0; while(l<=r)if(by[mid=(l+r)>>1]<=x)l=(t=mid)+1;else r=mid-1; return t; } int ins(int x,int a,int b,int c){ int y=++tot; v[y]=v[x]+p; if(a==b)return y; int mid=(a+b)>>1; if(c<=mid)l[y]=ins(l[x],a,mid,c),r[y]=r[x];else l[y]=l[x],r[y]=ins(r[x],mid+1,b,c); return y; } ll asko(int x,int a,int b,int d){ if(!x||!d)return 0; if(b<=d)return v[x].o; int mid=(a+b)>>1; ll t=asko(l[x],a,mid,d); if(d>mid)t+=asko(r[x],mid+1,b,d); return t; } ll askj(int x,int a,int b,int d){ if(!x||!d)return 0; if(b<=d)return v[x].j; int mid=(a+b)>>1; ll t=askj(l[x],a,mid,d); if(d>mid)t+=askj(r[x],mid+1,b,d); return t; } ll aski(int x,int a,int b,int d){ if(!x||!d)return 0; if(b<=d)return v[x].i; int mid=(a+b)>>1; ll t=aski(l[x],a,mid,d); if(d>mid)t+=aski(r[x],mid+1,b,d); return t; } ll askij(int x,int a,int b,int d){ if(!x||!d)return 0; if(b<=d)return v[x].ij; int mid=(a+b)>>1; ll t=askij(l[x],a,mid,d); if(d>mid)t+=askij(r[x],mid+1,b,d); return t; } inline ll sum(int x,int y){ int i=lowerx(x),j=lowery(y); return asko(T[i],1,cl,j)*(x+1)*(y+1)-askj(T[i],1,cl,j)*(x+1)-aski(T[i],1,cl,j)*(y+1)+askij(T[i],1,cl,j); } inline void read(int&a){char c;while(!(((c=getchar())>='0')&&(c<='9')));a=c-'0';while(((c=getchar())>='0')&&(c<='9'))(a*=10)+=c-'0';} int main(){ read(n),read(m),read(k),read(q); while(k--){ read(X1),read(X2),read(Y1),read(Y2),read(s); a[++cnt]=P(X1,Y1,s); a[++cnt]=P(X2+1,Y1,-s); a[++cnt]=P(X1,Y2+1,-s); a[++cnt]=P(X2+1,Y2+1,s); bx[++cl]=X1,by[cl]=Y1,bx[++cl]=X2+1,by[cl]=Y2+1; } sort(bx+1,bx+cl+1),sort(by+1,by+cl+1); for(i=1;i<=cnt;i++)nxt[i]=g[a[i].x=lowerx(a[i].x)],g[a[i].x]=i; for(i=1;i<=cl;i++)for(T[i]=T[i-1],j=g[i];j;j=nxt[j])p=V(bx[i],a[j].y,a[j].z),T[i]=ins(T[i],1,cl,lowery(a[j].y)); while(q--){ read(x),read(y); X1=ans%n+1,X2=(ans+x)%n+1;if(X1>X2)swap(X1,X2); Y1=ans%m+1,Y2=(ans+y)%m+1;if(Y1>Y2)swap(Y1,Y2); printf("%lld\n",ans=sum(X2,Y2)-sum(X1-1,Y2)-sum(X2,Y1-1)+sum(X1-1,Y1-1)); } return 0; }