LOJ 6278 数列分块入门2

【题解】

　　分块。块内排序。块内二分出第一个大于等于c的数。

　　

#include<cstdio>
#include<algorithm>
#include<cmath>
#define N 500010
#define rg register
using namespace std;
int n,m,opt,l,r,c,block,bl[N],tag[N];
struct rec{
	int data,pos;
}a[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;
}
inline bool cmp(rec x,rec y){return x.data<y.data;}
inline int find(int pos,int x){
	int l=(pos-1)*block+1,r=pos*block, ll=l;
	while(l<r){
		int mid=(l+r+1)>>1;
		if(a[mid].data<x) l=mid;
		else r=mid-1;
	}
	if(a[l].data>=x) return 0;
	return l-ll+1;
}
int main(){
	n=read(); block=sqrt(n);
	for(rg int i=1;i<=n;i++) a[i].data=read(),a[i].pos=i;
	for(rg int i=1;i<=n;i++) bl[i]=(i-1)/block+1;
	for(rg int i=1;i<=block+1;i++) sort(a+(i-1)*block+1,a+min(i*block,n)+1,cmp);
	for(rg int j=1;j<=n;j++){
		if((opt=read())==0){
			l=read(); r=read(); c=read();
			int st=bl[l]+1,ed=bl[r]-1;
			for(rg int i=st;i<=ed;i++) tag[i]+=c;
			if(bl[l]==bl[r]){
				int x=(bl[l]-1)*block+1,y=min(bl[l]*block,n);
				for(rg int i=x;i<=y;i++) if(a[i].pos>=l&&a[i].pos<=r) a[i].data+=c;
				sort(a+x,a+y+1,cmp);
				continue;
			}
			int x=(bl[l]-1)*block+1,y=min(bl[l]*block,n);
			for(rg int i=x;i<=y;i++)
				if(a[i].pos>=l) a[i].data+=c;
			sort(a+x,a+y+1,cmp);
			x=(bl[r]-1)*block+1,y=min(bl[r]*block,n);
			for(rg int i=x;i<=y;i++)
				if(a[i].pos<=r) a[i].data+=c;
			sort(a+x,a+y+1,cmp);
		}
		else{
			l=read(); r=read(); c=read(); c*=c;
			int st=bl[l]+1,ed=bl[r]-1,ans=0;
			for(rg int i=st;i<=ed;i++) ans+=find(i,c-tag[i]);
			if(bl[l]==bl[r]){
				for(rg int i=(bl[l]-1)*block+1;i<=bl[l]*block;i++)
					if(a[i].pos>=l&&a[i].pos<=r&&a[i].data+tag[bl[l]]<c) ans++;
				printf("%d\n",ans); continue;
			}
			int x=(bl[l]-1)*block+1,y=min(bl[l]*block,n);
			for(rg int i=x;i<=y;i++)
				if(a[i].pos>=l&&a[i].data+tag[bl[l]]<c) ans++;
			x=(bl[r]-1)*block+1,y=min(bl[r]*block,n);
			for(rg int i=x;i<=y;i++)
				if(a[i].pos<=r&&a[i].data+tag[bl[r]]<c) ans++;
			printf("%d\n",ans);
		}
	}
	return 0;
}