分块笔记

分块

分块1

单点查值，区间加值

#include <iostream>
#include <cmath>
using namespace std;
#define int long long
const int N = 5e4+10,M = 1000;
int a[N],n,m,ed[M],st[M],belong[N],add[M];
inline int read(){
    int res = 0, flag = 1;
    char ch = getchar();
    while(ch < '0'|| ch > '9'){
        if(ch == '-')flag = -1;
        ch = getchar();
    }
    while(ch >= '0'&& ch <= '9'){
        res = (res<<1) + (res<<3) + (ch^48);
        ch = getchar();
    }
    return res * flag;
}

void build(int n)
{
    int num = sqrt(n);
    for(int i = 1 ; i <= num ; i ++ )
    {
        st[i] = n / num * (i - 1) + 1;
        ed[i] = n / num * i;
    }
    ed[num] = n;
    for(int i = 1 ; i <= num ; i ++ )
        for(int j = st[i] ; j <= ed[i] ; j ++ )
        belong[j] = i;
}

void change(int l,int r,int c)
{
    int x = belong[l],y = belong[r];
    if(x == y)
    {
        for(int i = l;i <= r;i++)a[i] += c;
        return;
    }
    for(int i = l ; i <= ed[x] ; i ++ )a[i] += c;
    for(int i = st[y] ; i <= r ; i ++)a[i] += c;
    for(int i = x + 1 ; i <= y - 1; i ++ )add[i] += c;
}

int ask(int r)
{
    int x = belong[r];
    return a[r] + add[x];
}


signed main()
{
    //freopen("a1.in","r",stdin);
    //freopen("a_2.out","w",stdout);
    n = read();
    for(int i = 1 ; i <= n ; i ++ )a[i] = read();
    build(n);
    for(int opt,l,r,c,i = 1 ; i <= n ; i ++ )
    {
        opt = read();
        l = read();
        r = read();
        c = read();
        if(opt){
            printf("%lld\n",ask(r));
        }
        else {
            change(l,r,c);
        }
    }

}

分块二

给出一个长为 \(n\) 的数列，以及 \(n\) 个操作，操作涉及区间加法，询问区间内小于某个值 \(x\) 的元素个数。

#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
#define IOin(x) freopen(#x,"r",stdin)
#define IOout(x) freopen(#x,"w",stdout)
#define int long long 
const int N = 5e4 + 10,M =1e3;
int n,a[N],belong[N],st[M],ed[M],add[M],num;
inline int read(){
    int res = 0, flag = 1;
    char ch = getchar();
    while(ch < '0'|| ch > '9'){
        if(ch == '-')flag = -1;
        ch = getchar();
    }
    while(ch >= '0'&& ch <= '9'){
        res = (res<<1) + (res<<3) + (ch^48);
        ch = getchar();
    }
    return res * flag;
}

void build(int n)
{
    num  = sqrt(n);
    for(int i = 1 ; i <= num ; i ++ )
    {
        st[i] = n / num * (i - 1) + 1;
        ed[i] = n / num * i ;
    }
    ed[num] = n;
    for(int i = 1 ; i <= num ; i ++ )
    {
        for(int j = st[i] ; j <= ed[i] ; j ++ )
        belong[j] = i;
    }
}

void change(int l,int r,int c)
{
    int x = belong[l],y = belong[r];
    if(x == y)
    {
        for(int i = l ; i <= r ; i ++ )a[i] += c;
        return ; 
    }
    for(int i = l ; i <= ed[x] ; i ++ )a[i] += c;
    for(int i = st[y] ; i <= r ; i ++ )a[i] += c;
    for(int i = x + 1 ; i <= y - 1  ; i ++ )add[i] += c;   

}

int ask(int l,int r,int c)
{
    int cnt = 0;
    int x = belong[l],y = belong[r];
    if(x == y)
    {
        for(int i = l ; i <= r ; i ++ )
        if(a[i] + add[x] < c)cnt ++ ;
        return  cnt;
    }

    for(int i = l ; i <= ed[x] ; i ++ )
    if(a[i] + add[x] < c)cnt ++ ;
    for(int i = st[y] ; i <= r ; i ++ )
    if(a[i] + add[y] < c)cnt ++ ;
    for(int i = x + 1 ; i <= y - 1  ; i ++ )
    {
        for(int j = st[i] ; j <= ed[i] ; j ++ )
            if(a[j] + add[i] < c)cnt ++ ;
    }
    return cnt;
    
}


signed main()
{
    //IOin(a1.in);
    //IOout(a11.out);
    n = read();
    for(int i = 1 ; i <= n ; i ++ ) a[i] = read();
    build(n);
    for(int i = 1 ; i <= n ; i ++ )
    {
        int opt,l,r,c;
        opt  = read();
        l = read();
        r = read();
        c = read();
        if(!opt){
            change(l,r,c);
        }else {
            printf("%d\n",ask(l,r,c*c));
        }
    }
}

分块3

给出一个长为 \(n\) 的数列，以及 \(n\) 个操作，操作涉及区间加法，询问区间内小于某个值 \(x\) 的前驱（比其小的最大元素）。

#include <iostream>
#include <algorithm>
#include <cmath>
#include <set>
using namespace std;
const int N = 1e6 + 10 , M = 1e3 + 10;

int a[N],llt[N],b[N],n,len;
set<int>s[M];



inline int read()
{
	int res = 0,flag = 1;
	char ch = getchar();
	while(ch < '0' || ch > '9')
	{
		if(ch == '-') flag = -1;
		ch = getchar();
	}
	while(ch >= '0' && ch <= '9')
	{
		res = (res<<1) + (res << 3) + (ch^48);
		ch = getchar();
	}
	return res * flag ;
}



void add(int l,int r,int c)
{
	int x = llt[l] , y = llt[r];
	for (int i = l ; i <= min(x * len , r) ; i ++ )
	{
		s[x].erase(a[i]);
		a[i] += c;
		s[x].insert(a[i]);
	}
	if(x != y)
	{
		for (int i = len * (y - 1) + 1 ; i <= r ; i ++ )
		{
			s[y].erase(a[i]);
			a[i] += c;
			s[y].insert(a[i]);
		}
	}
	for (int i = x + 1 ; i <= y - 1 ; i ++ )
		b[i] += c;
}


void ask(int l,int r,int c)
{
	int ans = - 1;
	int x = llt[l], y = llt[r];
	for (int i = l ; i <= min(x * len ,  r) ; i ++ )
		if(a[i] + b[x] < c)
			ans = max(ans, a[i] + b[x]);
			
	if(x != y)
	{
		for (int i = len * (y - 1) + 1 ; i <= r  ; i ++ )
			if(a[i] + b[y] < c)
				ans = max(ans , a[i] + b[y]);
	}
	
	for (int i = x + 1 ; i <= y - 1 ; i ++ )
	{
		int t = c - b[i] ;
		set<int>::iterator it = s[i].lower_bound(t);
		if(it == s[i].begin()) continue;
		it -- ;
		ans = max(ans , *it + b[i]);
	}
	printf("%d\n",ans);
	
}


int main()
{
	n = read();
	len = 1000 ;
	for (int i = 1 ; i <= n ; i ++ )
	a[i] = read(),
	llt[i] = (i - 1) / len + 1,
	s[llt[i]].insert(a[i]);
	for (int i = 1 ; i <= n ; i ++ )
	{
		int opt , l , r , c;
		opt = read();
		l = read(); 
		r = read();
		c = read();
		if(opt == 1)
		ask(l,r,c);
		else
		add(l,r,c);
	}
	return 0;
	
}