【模板】分块

　　分块。利用根号平衡保证复杂度。修改跨过一整块的直接打标记，不足一整块的暴力修改。复杂度为$O(n\sqrt{n})$

　　LOJ 6277

#include<cstdio>
#include<algorithm>
#include<cmath>
#define N (500010)
#define rg register
using namespace std;
int n,a[N],bl[N],tag[N];
inline int read(){
	int k=0,f=1; char c=getchar();
	while(c<'0'||c>'9')c=='-'&&(f=-1),c=getchar();
	while('0'<=c&&c<='9')k=k*10+c-'0',c=getchar();
	return k*f;
}
int main(){
	n=read(); int block=sqrt(n);
	for(rg int i=1;i<=n;i++) a[i]=read(),bl[i]=(i-1)/block+1;
	for(rg int i=1;i<=n;i++){
		int opt=read(),l=read(),r=read(),del=read();
		if(!opt){
			int st=bl[l]+((l-1)%block?1:0),ed=bl[r]-(r%block?1:0);
			for(rg int j=st;j<=ed;j++) tag[j]+=del;
			if(bl[l]==bl[r]&&(r-l+1)<block){
				for(rg int j=l;j<=r;j++) a[j]+=del;
				continue;
			}
			if(st!=bl[l]) for(rg int j=l;j<=bl[l]*block;j++) a[j]+=del;
			if(ed!=bl[r]) for(rg int j=ed*block+1;j<=r;j++) a[j]+=del;
		}
		else printf("%d\n",a[r]+tag[bl[r]]);
	}
	return 0;
}