#6277. 数列分块入门 1

Archie

区间加和单点查询

很简单的思路就是\(O(\sqrt{n})修改和o(1)\)查询，就像线段树一样搞。一个tag

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
int n;
int be[50005];
int len;
int x,y,z,k;
int a[50004];
int tag[50004];
void add(int l,int r,int c){
	for(int i=l;i<=min(r,be[l]*len);++i){
		a[i]+=c;
	}
	if(be[l]!=be[r]){
		for(int i=(be[r]-1)*len+1;i<=r;++i){
			a[i]+=c;
		}
	}
	for(int i=be[l]+1;i<be[r];++i){
		tag[i]+=c;
	}
	return ;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
		scanf("%d",&a[i]);
	len=sqrt(n);
	for(int i=1;i<=n;++i){
		be[i]=(i-1)/len+1;
	}
	for(int i=1;i<=n;++i){
		scanf("%d%d%d%d",&x,&y,&z,&k);
		if(x==0){
			add(y,z,k);
		}else{
			cout<<a[z]+tag[be[z]]<<endl;;
		}
		
	}
	return 0;
}

或者说差分，变成单点修改区间查

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
int n;
int a[100005];
int be[100006];
int len;
int cha[1000006];
int sum[100006];
int k,x,y,z;
void xiu(int x,int c){
	cha[x]+=c;
	sum[be[x]]+=c;
}
int query(int l,int r){
	int ans=0;
	if(be[l]==be[r]){
		for(int i=l;i<=r;++i){
			ans+=cha[i];		
		}
	}else{
		for(int i=1;i<be[r];++i){
			ans+=sum[i];
		}
		for(int i=(be[r]-1)*len+1;i<=r;++i){
			ans+=cha[i];
		}
	}
	return ans;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i){
		scanf("%d",&a[i]);
		cha[i]=a[i]-a[i-1];
	}
	len=sqrt(n);
	for(int i=1;i<=n;++i){
		be[i]=(i-1)/len+1;
		sum[be[i]]+=cha[i];
	}
	for(int i=1;i<=n;++i){
		scanf("%d%d%d%d",&k,&x,&y,&z);
		if(k==0){
			xiu(x,z);
			xiu(y+1,-z);
		}else{
			cout<<query(1,y)<<endl;
		}
	}
	return 0;
}

标签：cha,50004,数列,6277,分块,int,sum,tag,include
来源： https://www.cnblogs.com/For-Miku/p/15022873.html