分析
考虑离线处理，那么询问区间和就可以转换为四个询问，
CDQ分治按横坐标处理询问，树状数组维护前缀和就可以了
代码
#include <cstdio>
#include <cctype>
#include <algorithm>
#define rr register
using namespace std;
const int N=2000011,M=200011;
struct rec{int x,X,Y;}q[M],Q[M];
int c[N],n,ans[M],CNT,TOT;
inline signed iut(){
rr int ans=0; rr char c=getchar();
while (!isdigit(c)) c=getchar();
while (isdigit(c)) ans=(ans<<3)+(ans<<1)+(c^48),c=getchar();
return ans;
}
inline void print(int ans){
if (ans>9) print(ans/10);
putchar(ans%10+48);
}
inline void add(int x,int y){for (;x<=n;x+=-x&x) c[x]+=y;}
inline signed query(int x){rr int ans=0; for (;x;x-=-x&x) ans+=c[x]; return ans;}
inline void CDQ(int l,int r){
if (l==r) return;
rr int mid=(l+r)>>1,i=l,j=mid+1,Cnt=0;
CDQ(l,mid),CDQ(mid+1,r);
for (;j<=r;++j){
for (;i<=mid&&q[i].X<=q[j].X;++i){
if (q[i].x>1e4) add(q[i].Y,q[i].x-1e4);
Q[++Cnt]=q[i];
}
if (q[j].x<=1e4){
if (q[j].x<0) ans[-q[j].x]-=query(q[j].Y);
else ans[q[j].x]+=query(q[j].Y);
}
Q[++Cnt]=q[j];
}
for (rr int I=l;I<i;++I)
if (q[I].x>1e4)
add(q[I].Y,1e4-q[I].x);
for (rr int I=i;I<=mid;++I) Q[++Cnt]=q[I];
for (rr int I=l;I<=r;++I) q[I]=Q[I-l+1];
}
signed main(){
iut(),n=iut();
for (rr int z=iut();z!=3;z=iut()){
if (z==1){
rr int x=iut(),y=iut(),w=iut();
q[++CNT]=(rec){w+1e4,x,y};
}else{
rr int lx=iut(),ly=iut(),rx=iut(),ry=iut();
++TOT,q[++CNT]=(rec){TOT,rx,ry};
if (lx>1&&ly>1) q[++CNT]=(rec){TOT,lx-1,ly-1};
if (lx>1) q[++CNT]=(rec){-TOT,lx-1,ry};
if (ly>1) q[++CNT]=(rec){-TOT,rx,ly-1};
}
}
CDQ(1,CNT);
for (rr int i=1;i<=TOT;++i)
print(ans[i]),putchar(10);
return 0;
}