hzwer的9题https://loj.ac/problem/6277https://loj.ac/problem/6278https://loj.ac/problem/6279https://loj.ac/problem/6280https://loj.ac/problem/6281https://loj.ac/problem/6282https://loj.ac/problem/6283https://loj.ac/problem/6284https://loj.
hzwer的9题 https://loj.ac/problem/6277 https://loj.ac/problem/6278 https://loj.ac/problem/6279 https://loj.ac/problem/6280 https://loj.ac/problem/6281 https://loj.ac/problem/6282 https://loj.ac/problem/6283 https://loj.ac/problem/6284 https://loj.ac/problem/6285
例题会稍后放上
分块是一种优雅的暴力
大概思想是这样的:维护\(\sqrt(n)\)个块 查询的时候是查询区间内的块 如果有的部分不满一个块就用枚举 常数较小
对于上面这个说明 我举个简单的例子： 假设n = 10000 那么 每个块就是\(\sqrt n\) = 100 每个块是
1~100 101~200 ... 9901~10000
假设要查询 2 ~ 999的值
你只需要维护每个 大小 为 \(\sqrt n\) 的块 也就是 大小 为 100 的块 然后对\(101\)~\(200\) - \(801\)~\(900\)这几个块查询块的值
然后暴力 \(2\)~\(100\) 和 \(901\) ~ \(999\) 的值
之所以称为优雅的暴力 分块把朴素暴力的 \(2\) ~ \(999\) 改成了 \(200\) 复杂度左右的暴力
个人觉得 \(2\)~\(999\)是最坏情况
如果查询 \(1\)~\(1000\) 那么分块的优势就出来了
分块的预处理是 将每一个数字存入一个块中 复杂度是\(\theta n\)
然后区间修改查询什么的 最多不超过 \(3\sqrt n\)
证明：一个块的大小是\(\sqrt(n)\) 那么多出来的左区间和右区间的最坏情况是 \(\sqrt(n)\)) 所以易证
分块的时间复杂度大概就是 \(\theta (N + q * \sqrt(N))\)
（n指序列长度 q指查询修改的操作次数）
以上就是一个基本的分块思想 简单讲 分块比线段树容易实现
\(\text{分块1}\)
分块1是区间修改 单点查询
我们用一个数组维护块
如果区间修改的时候 包含这个块 那么可以把这个块加上需要修改的值 如果是多出来的就直接暴力修改了
然后查询的时候输出所在块修改的值 和 本身的值
// Isaunoya
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace std ;
const int N = 50000 + 5 ;
const int Bl = 300 + 5 ;
struct node {
int l , r ;
int add ;
} ;
int n ;
int a[N] ;
node atag[Bl] ;
int bl[N] ; int unt ;
inline void change(int l , int r , int c) {
for(register int i = l ; i <= min(bl[l] * unt , r) ; i ++)
a[i] += c ;
if(bl[l] != bl[r])
for(register int i = (bl[r] - 1) * unt + 1 ; i <= r ; i ++)
a[i] += c ;
for(register int i = bl[l] + 1 ; i <= bl[r] - 1 ; i ++)
atag[i].add += c ;
}
signed main() {
ios::sync_with_stdio(false) ;
cin >> n ;
for(register int i = 1 ; i <= n ; i ++) cin >> a[i] ;
unt = sqrt(n) ;
for(register int i = 1 ; i <= n ; i ++) {
bl[i] = (i - 1) / unt + 1 ;
}
for(register int i = 1 ; i <= n ; i ++) {
int opt ;
cin >> opt ;
if(opt == 0) {
int l , r , c ;
cin >> l >> r >> c ;
change(l , r , c) ;
}
else {
int l , r , c ;
cin >> l >> r >> c ;
cout << a[r] + atag[bl[r]].add << endl ;
}
}
return 0 ;
}
\(\text{分块2}\)
分块二求的是区间修改 区间查询小于\(c^2\)的最大数
vector 每次修改完重构一下左右块就行了
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
inline int read() {
register int x = 0;
register int f = 1;
register char c;
#define gc c = getchar()
while (isspace(gc))
;
c == '-' ? gc, f = -1 : 0;
while (x = (x << 3) + (x << 1) + (c & 15), isdigit(gc))
;
return x * f;
}
const int N = 50000 + 5;
const int Bl = 300 + 5;
int n;
int a[N];
struct node {
int add;
std::vector<int> v;
};
node atag[Bl];
int unt;
int bl[N];
inline void reset(int x) {
atag[x].v.clear();
for (register int i = (x - 1) * unt + 1; i <= min(x * unt, n); i++) atag[x].v.push_back(a[i]);
sort(atag[x].v.begin(), atag[x].v.end());
return;
}
inline void change(int l, int r, int c) {
for (register int i = l; i <= min(bl[l] * unt, r); i++) a[i] += c;
reset(bl[l]);
if (bl[l] != bl[r]) {
for (register int i = (bl[r] - 1) * unt + 1; i <= r; i++) a[i] += c;
reset(bl[r]);
}
for (register int i = bl[l] + 1; i <= bl[r] - 1; i++) atag[i].add += c;
}
inline int query(int l, int r, LL c) {
int ans = 0;
for (register int i = l; i <= min(bl[l] * unt, r); i++)
if (a[i] + atag[bl[l]].add < c)
ans++;
if (bl[l] != bl[r]) {
for (register int i = (bl[r] - 1) * unt + 1; i <= r; i++)
if (a[i] + atag[bl[r]].add < c)
ans++;
}
for (register int i = bl[l] + 1; i <= bl[r] - 1; i++) {
int s = c - atag[i].add;
ans += lower_bound(atag[i].v.begin(), atag[i].v.end(), s) - atag[i].v.begin();
}
return ans;
}
signed main() {
n = read();
unt = sqrt(n);
for (register int i = 1; i <= n; i++) a[i] = read();
for (register int i = 1; i <= n; i++) bl[i] = (i - 1) / unt + 1;
for (register int i = 1; i <= n; i++) {
atag[bl[i]].v.push_back(a[i]);
}
for (register int i = 1; i <= bl[n]; i++) {
sort(atag[i].v.begin(), atag[i].v.end());
}
for (register int i = 1; i <= n; i++) {
int opt = read(), L = read(), R = read(), c = read();
if (opt)
printf("%lld\n", query(L, R, c * c));
else
change(L, R, c);
}
return 0;
}
\(\text{分块3}\)
分块3 区间修改 询问区间内小于某个值 x的前驱即比x小的最大数
同样使用一个set维护
然后用\(lower _ \ bound\)二分
（部分C++11内容
// Isaunoya
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC diagnostic error "-std=c++11"
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
inline int read() {
register int x = 0;
register int f = 1;
register char c;
#define gc c = getchar()
while (isspace(gc))
;
c == '-' ? gc, f = -1 : 0;
while (x = (x << 3) + (x << 1) + (c & 15), isdigit(gc))
;
return x * f;
}
const int N = 100000 + 5;
const int Bl = 400 + 5;
int n;
int a[N];
struct node {
int add;
};
node atag[Bl];
int unt;
int bl[N];
set<int> st[Bl];
inline void change(int l, int r, int c) {
for (register int i = l; i <= min(bl[l] * unt, r); i++) {
st[bl[i]].erase(a[i]);
a[i] += c;
st[bl[i]].insert(a[i]);
}
if (bl[l] != bl[r]) {
for (register int i = (bl[r] - 1) * unt + 1; i <= r; i++) {
st[bl[i]].erase(a[i]);
a[i] += c;
st[bl[i]].insert(a[i]);
}
}
for (register int i = bl[l] + 1; i <= bl[r] - 1; i++) atag[i].add += c;
}
inline int query(int l, int r, int c) {
int ans = -1;
for (register int i = l; i <= min(bl[l] * unt, r); i++)
if (a[i] + atag[bl[l]].add < c)
ans = max(a[i] + atag[bl[l]].add, ans);
if (bl[l] != bl[r]) {
for (register int i = (bl[r] - 1) * unt + 1; i <= r; i++)
if (a[i] + atag[bl[r]].add < c)
ans = max(a[i] + atag[bl[l]].add, ans);
}
for (register int i = bl[l] + 1; i <= bl[r] - 1; i++) {
int s = c - atag[i].add;
auto find = st[i].lower_bound(s);
if (find == st[i].begin()) continue ;
find--;
ans = max(ans, *find + atag[i].add);
}
return ans;
}
signed main() {
n = read();
unt = sqrt(n);
for (register int i = 1; i <= n; i++) a[i] = read();
for (register int i = 1; i <= n; i++) bl[i] = (i - 1) / unt + 1;
for (register int i = 1; i <= n; i++) {
st[bl[i]].insert(a[i]);
}
for (register int i = 1; i <= n; i++) {
int opt = read(), L = read(), R = read(), c = read();
if (opt)
printf("%d\n", query(L, R, c));
else
change(L, R, c);
}
return 0;
}
\({\text{分块4}}\)
分块4是区间修改 区间求和
然后 区间和 % \(c+1\)
我们考虑使用一个数组维护区间和 然后用一个数组维护整个块的修改情况
维护区间和指的是 暴力修改 的部分
// Isaunoya
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace std ;
#define int long long
inline int read() { register int x = 0 ; register int f = 1 ; register char c ;
#define gc c = getchar()
while(isspace(gc)) ;
c == '-' ? gc , f = -1 : 0 ;
while(x = (x << 1) + (x << 3) + (c ^ 48) , isdigit(gc)) ;
return x * f ;
}
int n ;
const int N = 50000 + 5 ;
int a[N] ;
int bl[N] ;
int sum[N] ;
int atag[N] ;
int unt ;
inline void change(int l , int r , int c) {
for(register int i = l ; i <= min(r , bl[l] * unt) ; i ++) {
a[i] += c ;
sum[bl[l]] += c ;
}
if(bl[l] != bl[r])
for(register int i = (bl[r] - 1) * unt + 1 ; i <= r ; i ++) {
a[i] += c ;
sum[bl[r]] += c ;
}
for(register int i = bl[l] + 1 ; i <= bl[r] - 1 ; i ++) atag[i] += c ;
return ;
}
inline int query(int l , int r , int c) { int ans = 0 ;
for(register int i = l ; i <= min(r , bl[l] * unt) ; i ++) {
ans += a[i] + atag[bl[l]] ;
ans %= c ;
}
if(bl[l] != bl[r])
for(register int i = (bl[r] - 1) * unt + 1 ; i <= r ; i ++) {
ans += a[i] + atag[bl[r]] ;
ans %= c ;
}
for(register int i = bl[l] + 1 ; i <= bl[r] - 1 ; i ++)
ans = (ans + sum[i] + atag[i] * unt) % c ;
return ans ;
}
signed main() {
n = read() ; unt = sqrt(n) ;
for(register int i = 1 ; i <= n ; i ++) a[i] = read() ;
for(register int i = 1 ; i <= n ; i ++) bl[i] = (i - 1) / unt + 1 ;
for(register int i = 1 ; i <= n ; i ++) {
sum[bl[i]] += a[i] ;
}
for(register int i = 1 ;i <= n ; i ++) {
int opt = read() ;
if(opt == 0) {
int l = read() , r = read() , c = read() ;
change(l , r , c) ;
}
if(opt == 1) {
int l = read() , r = read() , c = read() ;
printf("%lld\n" , query(l , r , c + 1)) ;
}
}
return 0 ;
}
\(\text{分块5}\)
分块5是区间开方 然后区间查询
对于每个块 大小最大是\(2^{31}\)次
\(log(31) ≈ 5\)
所以可以得出一个块最多开方6次
也就是最大是\(6n\)
所以对于区间开方 对一个块进行开方 然后重构这个块的总和
也就是可以得出 大块修改并用一个flg数组标记这个块有没有大于1的数字（如果大于1的话还可以开方
这样可以避免很多次重复开方
那么对于没有完整块的左右区间 我们考虑先减掉原数字然后加上开方后的数字（对于乘法操作也是一样的
区间查询上面讲过了（将块的总和加上 然后暴力把左右区间求和）
// Isaunoya
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace std ;
#define int long long
inline int read() { register int x = 0 ; register int f = 1 ; register char c ;
#define gc c = getchar()
while(isspace(gc)) ;
c == '-' ? gc , f = -1 : 0 ;
while(x = (x << 1) + (x << 3) + (c ^ 48) , isdigit(gc)) ;
return x * f ;
}
int n ;
const int N = 50000 + 5 ;
int a[N] ;
int bl[N] ;
int flg[N] ;
int sum[N] ;
int unt ;
inline void Reset(int x) {
if(flg[x]) return ;
flg[x] = 1 ;
sum[x] = 0 ;
for(register int i = (x - 1) * unt + 1 ; i <= x * unt ; i ++) {
a[i] = sqrt(a[i]) ;
sum[x] += a[i] ;
if(a[i] > 1) flg[x] = 0 ;
}
return ;
}
inline void change(int l , int r , int c) {
for(register int i = l ; i <= min(bl[l] * unt , r) ; i ++) {
sum[bl[l]] -= a[i] ;
a[i] = sqrt(a[i]) ;
sum[bl[l]] += a[i] ;
}
if(bl[l] != bl[r])
for(register int i = (bl[r] - 1) * unt + 1 ; i <= r ; i ++) {
sum[bl[r]] -= a[i] ;
a[i] = sqrt(a[i]) ;
sum[bl[r]] += a[i] ;
}
for(register int i = bl[l] + 1 ; i <= bl[r] - 1 ; i ++) Reset(i) ;
return ;
}
inline int query(int l , int r , int c) { int ans = 0 ;
for(register int i = l ; i <= min(bl[l] * unt , r) ; i ++) {
ans += a[i] ;
}
if(bl[l] != bl[r])
for(register int i = (bl[r] - 1) * unt + 1 ; i <= r ; i ++) ans += a[i] ;
for(register int i = bl[l] + 1 ; i <= bl[r] - 1 ; i ++) ans += sum[i] ;
return ans ;
}
signed main() {
n = read() ; unt = sqrt(n) ;
for(register int i = 1 ; i <= n ; i ++) a[i] = read() ;
for(register int i = 1 ; i <= n ; i ++) bl[i] = (i - 1) / unt + 1 ;
for(register int i = 1 ; i <= n ; i ++) { sum[bl[i]] += a[i] ; }
for(register int i = 1 ; i <= n ; i ++) {
int opt = read() ;
if(opt == 0) {
int l , r , c ;
l = read() ; r = read() ; c = read() ;
change(l , r , c) ;
}
if(opt == 1) {
int l , r , c ;
l = read() ; r = read() ; c = read() ;
printf("%lld\n" , query(l , r , c)) ;
}
}
return 0 ;
}
\(\text{分块6}\)
分块6是可以区间插入一个数字 然后查询某个位置
（其实用不到分块 \(vector\) 可以直接做）
只是因为这一题没有区间查询 所以可以 \(vector\) 水过去
如果想好好学分块就使用分块的做法
当某个块大于 20 个块的时候 就重构块
不然查询的时候比较费劲 而且暴力复杂度比较高
// Isaunoya
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC diagnostic error "-std=c++11"
#include<bits/stdc++.h>
using namespace std ;
#define int long long
inline int read() { register int x = 0 ; register int f = 1 ; register char c ;
#define gc c = getchar()
while(isspace(gc)) ;
c == '-' ? gc , f = -1 : 0 ;
while(x = (x << 1) + (x << 3) + (c ^ 48) , isdigit(gc)) ;
return x * f ;
}
int n ;
int unt ;
const int N = 200000 + 5 ;
int a[N] ;
std::vector< int > v[N] ;
int st[N] ;
int top = 0 ; int m = 0 ;
pair < int , int > query(int b) {
int x = 1 ;
for ( ; b > v[x].size() ; ) b -= v[x ++].size() ;
return make_pair(x , b - 1) ;
}
inline void Rebuild() {
top = 0 ;
for(register int i = 1 ; i <= m ; i ++) {
for(auto j : v[i]) st[++ top] = j ;
v[i].clear() ;
}
int blo = sqrt(top) ;
for(register int i = 1 ; i <= top ; i ++) {
v[(i - 1) / blo + 1].push_back(st[i]) ;
}
m = (top - 1) / blo + 1 ;
}
inline void Insert(int x , int y) {
pair < int , int > p = query(x) ;
v[p.first].insert(v[p.first].begin() + p.second , y) ;
if(v[p.first].size() > 20 * unt) Rebuild() ;
}
signed main() {
n = read() ; unt = sqrt(n) ;
for(register int i = 1 ; i <= n ; i ++) a[i] = read() ;
for(register int i = 1 ; i <= n ; i ++) v[(i - 1) / unt + 1].push_back(a[i]) ;
m = (n - 1) / unt + 1 ;
for(register int i = 1 ; i <= n ; i ++) {
int opt = read() ;
if(opt == 0) {
int l = read() , r = read() , c = read() ;
Insert(l , r) ;
}
if(opt == 1) {
int l = read() , r = read() , c = read() ;
pair < int , int > p = query(r) ;
printf("%d\n" , v[p.first][p.second]) ;
}
}
return 0 ;
}
\(\text{分块7}\)
分块7是一题 \(\text{区间乘法 区间加法 单点查询}\) 的题目
那么我们只需要维护一下区间的值 乘法的值 加法的值就可以
如果乘法那么就需要把加法的值乘上一个值
不过在每次修改的时候 需要\(\text{重构这个不完整的块所在的块}\)
（好拗口 反正就是在\(左，右区间\)所在的块 然后把乘的值归1 然后加法的值清0
这样的话复杂度仍然还是 \(\sqrt n\)
根据这个思路的话 我写了一个区间查询的
我告诉我旁边的人：知道什么叫做单点查询吗
-- \(query(b,b)\)
// Isaunoya
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC diagnostic error "-std=c++11"
#include<bits/stdc++.h>
using namespace std ;
#define int long long
#define rep(i , j , n) for(register int i=j;i<=n;i++)
#define Rep(i , j , n) for(register int i=j;i>=n;i--)
#define gc c = getchar()
#define int long long
inline int read() { register int x = 0 , f = 1 ; register char c ;
while(isspace(gc)) ; c == '-' ? f = -1 , gc : 0 ;
while((x *= 10) += (c ^ 48) , isdigit(gc)) ;
return x * f ;
}
using namespace std ;
int n ;
const int N = 100000 + 5 ;
const int Unt = 2000 ;
const int p = 10000 + 7 ;
int a[N] ;
int unt ;
int block[N] ;
int sum[Unt] ;
int mul[Unt] ;
int size[Unt] ;
int ans[Unt] ;
inline void reset(int x) {
for(register int i=(x - 1) * unt + 1 ; i <= min(n , x * unt) ; i ++) {
a[i] = (a[i] * mul[x] + sum[x]) % p ;
}
sum[x] = 0 ; mul[x] = 1 ;
}
inline void change_mul(int l , int r , int x) {
reset(block[l]) ;
for(register int i=l;i<=min(r , block[l] * unt) ; i ++) {
ans[block[l]] -= a[i] ;
a[i] *= x ;
a[i] %= p ;
ans[block[l]] += a[i] ;
ans[block[l]] %= p ;
}
if(block[l] != block[r]) {
reset(block[r]) ;
for(register int i=(block[r] - 1) * unt + 1 ; i <= r ; i ++) {
ans[block[r]] -= a[i] ;
a[i] *= x ;
a[i] %= p ;
ans[block[r]] += a[i] ;
ans[block[r]] %= p ;
}
for(register int i=block[l] + 1 ; i <= block[r] - 1 ; i ++) {
sum[i] *= x ;
sum[i] %= p ;
mul[i] *= x ;
mul[i] %= p ;
ans[i] *= x ;
ans[i] %= p ;
}
}
}
inline void change_sum(int l , int r , int x) {
reset(block[l]) ;
for(register int i=l ; i <= min(r , block[l] * unt) ; i ++) {
a[i] += x ;
a[i] %= p ;
ans[block[l]] += x ;
}
if(block[l] != block[r]) {
reset(block[r]) ;
for(register int i = (block[r] - 1) * unt + 1 ; i <= r ; i ++) {
a[i] += x ;
a[i] %= p ;
ans[block[r]] += x ;
}
for(register int i=block[l] + 1 ; i <= block[r] - 1 ; i ++) {
sum[i] += x ;
sum[i] %= p ;
ans[i] += x * size[i] ;
}
}
}
inline int Query(int l , int r) {
int Ans = 0 ;
for(register int i=l;i<=min(r , block[l] * unt) ; i ++)
Ans += (a[i] * mul[block[l]] + sum[block[l]] ) % p ;
if(block[l] != block[r]) {
for(register int i = (block[r] - 1) * unt + 1 ; i <= r ; i ++)
Ans += (a[i] * mul[block[r]] + sum[block[r]] ) % p ;
for(register int i = block[l] + 1 ; i <= block[r] - 1 ; i ++)
Ans += ans[i] ;
}
return Ans % p ;
}
signed main() {
n = read() ;
unt = sqrt(n) ;
for(register int i=1;i<=n;i++) a[i] = read() ;
for(register int i=1;i<=n;i++) {
block[i] = (i - 1) / unt + 1 ;
ans[block[i]] += a[i] ;
size[block[i]] ++ ;
}
for(register int i=1;i<=block[n];i++) mul[i] = 1 ;
for(register int i=1;i<=n;i++) {
int opt = read() ;
if(opt == 0) {
int a = read() , b = read() , c = read() ;
change_sum(a , b , c) ;
}
if(opt == 1) {
int a = read() , b = read() , c = read() ;
change_mul(a , b , c) ;
}
if(opt == 2) {
int a = read() , b = read() , c = read() ;
printf("%lld\n" , Query(b , b)) ;
}
}
return 0 ;
}
\(\text{分块8}\)
分块8的话是一个 查询区间有多少个\(c\) 并把整个区间改成 \(c\)
同样我们考虑判重 把每个块用一个\(flg\)数组记录当前值
然后对于左右区间的修改 把左右区间所在的块重构成之前的\(flg\)
然后修改成c 再判断左右区间有多少个c
那么对于\(\text{完整的块}\) 我们只需要 加上一个块的长度
这样就可以做到判重的效果了
对于 \(\text{完整的块}\) 我们只需要修改当前块的\(flg\)值就可以了
// Isaunoya
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC diagnostic error "-std=c++11"
#include<bits/stdc++.h>
using namespace std ;
#define int long long
#define rep(i , j , n) for(register int i=j;i<=n;i++)
#define Rep(i , j , n) for(register int i=j;i>=n;i--)
#define gc c = getchar()
#define int long long
inline int read() { register int x = 0 , f = 1 ; register char c ;
while(isspace(gc)) ; c == '-' ? f = -1 , gc : 0 ;
while((x *= 10) += (c ^ 48) , isdigit(gc)) ;
return x * f ;
}
const int N = 100000 + 5 ;
int n ;
int a[N] ;
int unt ;
int flg[N] ;
int bl[N] ;
inline void reset(int x) {
if(flg[x] == -1) return ;
for(register int i = (x - 1) * unt + 1 ; i <= x * unt ; i ++) a[i] = flg[x] ;
flg[x] = -1 ;
}
inline int solve(int l , int r , int c) { int ans = 0 ;
reset(bl[l]) ;
for(register int i = l ; i <= min(bl[l] * unt , r) ; i ++) {
if(a[i] != c) a[i] = c ;
else ans ++ ;
}
if(bl[l] != bl[r]) {
reset(bl[r]) ;
for(register int i = (bl[r] - 1) * unt + 1 ; i <= r ; i ++) {
if(a[i] != c) a[i] = c ;
else ans ++ ;
}
}
for(register int i = bl[l] + 1 ; i <= bl[r] - 1 ; i ++) {
if(flg[i] != -1) {
if(flg[i] != c) flg[i] = c ;
else ans += unt ;
}
else {
for(register int j = (i - 1) * unt + 1 ; j <= i * unt ; j ++)
if(a[j] != c) a[j] = c ;
else ans ++ ;
flg[i] = c ;
}
}
return ans ;
}
signed main() {
n = read() ; unt = sqrt(n) ;
for(register int i = 1 ; i <= n ; i ++) a[i] = read() ;
for(register int i = 1 ; i <= n ; i ++) bl[i] = (i - 1) / unt + 1 ;
for(register int i = 1 ; i <= bl[n] ; i ++) flg[i] = -1 ;
for(register int i = 1 ; i <= n ; i ++) {
int l = read() , r = read() , c = read() ;
printf("%lld\n" , solve(l , r , c)) ;
}
return 0 ;
}
\(\text{分块9}\)
前置知识：离线莫队
分块9的话我没想到怎么在线做
就是按每次查询的 \(l\) 排序 然后瞎搞
（大概是一个莫队的思想
// Isaunoya
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC diagnostic error "-std=c++11"
#include <bits/stdc++.h>
using namespace std;
#define rep(i, j, n) for (register int i = j; i <= n; i++)
#define Inc(i, j, n) for (register int i = j; i <= n; i++)
#define Rep(i, j, n) for (register int i = j; i >= n; i--)
inline int read() {
register int x = 0, f = 1;
register char c;
#define gc c = getchar()
while (isspace(gc))
;
c == '-' ? f = -1, gc : 0;
while ((x *= 10) += (c ^ 48), isdigit(gc))
;
return x * f;
}
const int N = 1e5;
int n, siz, k[N + 10], bl[N + 10];
struct Que {
int L, r, idx;
} q[N + 10];
bool cmp(const Que &A, const Que &B) {
return bl[A.L] ^ bl[B.L] ? A.L < B.L : A.r < B.r;
}
inline void init() {
n = read() ;
Inc(i, 1, n) k[i] = read() ;
Inc(i, 1, n) {
int x, y;
x = read() , y = read() ;
q[i] = (Que) {
x, y, i
};
}
siz = sqrt(n);
Inc(i, 1, n) bl[i] = (i - 1) / siz + 1;
sort(q + 1, q + 1 + n, cmp);
}
int rp[N + 10];
inline void disc() {
int tmp[N + 10];
Inc(i, 1, n) tmp[i] = k[i];
sort(tmp + 1, tmp + 1 + n);
int len = unique(tmp + 1, tmp + 1 + n) - tmp - 1;
Inc(i, 1, n) rp[i] = lower_bound(tmp + 1, tmp + 1 + len, k[i]) - tmp;
}
int cur_ans, cur_num, Ans[N + 10], num[N + 10];
inline void addr(int x) {
++num[rp[x]];
if (num[rp[x]] >= cur_num) {
if (num[rp[x]] == cur_num && k[x] < cur_ans)
cur_ans = k[x];
else if (num[rp[x]] > cur_num)
cur_ans = k[x], cur_num = num[rp[x]];
}
}
inline void addl(int x, int &ans, int &Num) {
++num[rp[x]];
if (num[rp[x]] >= Num) {
if (num[rp[x]] == Num && k[x] < ans)
ans = k[x];
else if (num[rp[x]] > Num)
ans = k[x], Num = num[rp[x]];
}
}
inline void remove(int x) {
--num[rp[x]];
}
inline void solv() {
int L, r, lim;
Inc(i, 1, n) {
if (bl[q[i].L] ^ bl[q[i - 1].L]) {
memset(num, 0, sizeof(num));
L = lim = bl[q[i].L] * siz + 1;
r = bl[q[i].L] * siz;
cur_ans = cur_num = 0;
}
if (bl[q[i].L] == bl[q[i].r]) {
int ans, nownum = 0;
Inc(j, q[i].L, q[i].r)++ num[rp[j]];
Inc(j, q[i].L, q[i].r) if (num[rp[j]] >= nownum) {
if (num[rp[j]] == nownum && k[j] < ans)
ans = k[j];
else if (num[rp[j]] > nownum)
ans = k[j], nownum = num[rp[j]];
}
Inc(j, q[i].L, q[i].r)-- num[rp[j]];
Ans[q[i].idx] = ans;
continue;
}
while (r < q[i].r) addr(++r);
int now = cur_ans, nownum = cur_num;
while (L > q[i].L) addl(--L, now, nownum);
Ans[q[i].idx] = now;
while (L < lim) remove(L++);
}
Inc(i, 1, n) printf("%d\n" , Ans[i]) ;
}
signed main() {
init();
disc();
solv();
return 0;
}
\(\text{例题}\)
https://loj.ac/problem/10117
\(\text{LOJ10117}\)
这题就是一个裸的树状数组 但是可以用分块做 好像更方便呢
// Isaunoya
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
#define rep(i , j , n) for(register int i = j ; i <= n ; i ++)
#define Rep(i , j , n) for(register int i = j ; i >= n ; i --)
#define to(u) for(register int i = head[u] ; i ; i = edge[i].nxt)
inline int read() { register int x = 0 , f = 1 ; register char c ;
#define gc c = getchar()
while(isspace(gc)) ; c == '-' ? f = -1 , gc : 0 ;
while(x = (x << 1) + (x << 3) + (c & 15) , isdigit(gc)) ;
return x * f ;
}
using namespace std ;
int n ;
const int N = 1e5 + 10 ;
int a[N] ;
int atag[N] ;
int bl[N] ;
int unt ;
inline void change(int l , int r) {
for(register int i = l ; i <= min(bl[l] * unt , r) ; i ++) a[i] ^= 1 ;
if(bl[l] != bl[r])
for(register int i = (bl[r] - 1) * unt + 1 ; i <= r ; i ++) a[i] ^= 1 ;
for(register int i = bl[l] + 1 ; i <= bl[r] - 1 ; i ++) atag[i] ^= 1 ;
return ;
}
signed main() {
// freopen(".in" , "r" , stdin) ; freopen(".out" , "w" , stdout) ;
n = read() ; unt = sqrt(n) ;
for(register int i = 1 ; i <= n ; i ++) bl[i] = (i - 1) / unt + 1 ;
for(register int q = read() ; q -- ; ) {
int opt = read() ;
if(opt & 1) {
int l = read() , r = read() ;
change(l , r) ;
}
else {
int x = read() ;
printf("%lld\n" , a[x] ^ atag[bl[x]]) ;
}
}
return 0 ;
}
https://www.luogu.org/problem/P3372 https://www.luogu.org/problem/P3373 https://www.luogu.org/problem/P2572 https://www.luogu.org/problem/P3396 https://www.luogu.org/problem/P3203 https://www.luogu.org/problem/P4879 https://www.luogu.org/problem/P3870 https://www.luogu.org/problem/P2574 https://www.luogu.org/problem/P2801 https://www.luogu.org/problem/P4145