Luogu P2801 Master's Magic Problem Solving

topic

When I first saw this question, I used a tree array to optimize the constant optimization of prefix and difference, and it turned out to be AC? (This data is too watery~)

The first block question I did, I debugged it for a long time, but I didn't expect that there would be an error on the second score! (must be noted) only this commemoration;

#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
int a[1000010],b[1000010],belong[1000010]; //原来的 
int lazy[1000010],l[1000010],r[1000010]; //块上的 
int block,tot,n,m;
void build()
{
	block=sqrt(n);
	tot=n/block;
	if(n%block){
		++tot;
	}
	for(register int i=1;i<=n;i++){
		b[i]=a[i];
		belong[i]=(i-1)/block+1;
	}
	for(register int i=1;i<=tot;i++){
		l[i]=(i-1)*block+1;
		r[i]=i*block;
	}
	r[tot]=n;
	for(register int i=1;i<=tot;i++){
		sort(b+l[i],b+r[i]+1);
	}
	return;
}
void change(int x,int y,int k)
{
	if(belong[x]==belong[y]){
		for(register int i=x;i<=y;i++){
			a[i]+=k;
		}
		for(register int i=l[belong[x]];i<=r[belong[x]];i++){
			b[i]=a[i];
		}
		sort(b+l[belong[x]],b+r[belong[x]]+1);
		return ;
	}
	for(register int i=x;i<=r[belong[x]];i++){
		a[i]+=k;
	}
	for(register int i=l[belong[x]];i<=r[belong[x]];i++){
		b[i]=a[i];
	}
	sort(b+l[belong[x]],b+r[belong[x]]+1);
	for(register int i=l[belong[y]];i<=y;i++){
		a[i]+=k;
	}
	for(register int i=l[belong[y]];i<=r[belong[y]];i++){
		b[i]=a[i];
	}
	sort(b+l[belong[y]],b+r[belong[y]]+1);
	for(register int i=belong[x]+1;i<=belong[y]-1;i++){
		lazy[i]+=k;
	}
}
int query(int x,int y,int goal)
{
	int ans=0;
	if(belong[x]==belong[y]){
		for(register int i=x;i<=y;i++){
			if(a[i]+lazy[belong[x]]>=goal) ++ans;
		}
		return ans;
	}
	for(register int i=x;i<=r[belong[x]];i++){
		if(a[i]+lazy[belong[x]]>=goal) ++ans;
	}
	for(register int i=l[belong[y]];i<=y;i++){
		if(a[i]+lazy[belong[y]]>=goal) ++ans;
	}
	for(register int i=belong[x]+1;i<=belong[y]-1;i++){
		int L=l[i],R=r[i],mid;
		while(L<R){
			int mid=(L+R)/2;
			if(b[mid]+lazy[i]>=goal){
				R=mid;
			}
			else{
				L=mid+1;
			}
		}
		if(L==r[i]){
			ans+=((b[L]+lazy[i])>=goal);
			continue;
		}
		ans+=(r[i]-L+1);
	}
	return ans;
} 
int main()
{
	cin>>n>>m;
	for(register int i=1;i<=n;i++) scanf("%d",&a[i]);
	build();
	for(register int i=1;i<=m;i++){
		char type;
		int x,y,k;
		cin>>type>>x>>y>>k;
		if(type=='M'){
			change(x,y,k);
		}
		else{
			cout<<query(x,y,k)<<endl;
		}
	}
} 

 

Reprinted in: https://www.cnblogs.com/kamimxr/p/11600198.html

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