数列分块入门1-9
分块1
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 50003;
int n, opt, a, b, c, cnt;
int w[maxn], in[maxn], addtag[maxn];
inline int read() {
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch <= '9' && ch >= '0') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
inline void add(int l, int r, int x) {
for (int i = l; i <= min(r, in[l]*cnt); i++)
w[i] += x;
if (in[l] != in[r])
for (int i = (in[r] - 1) * cnt + 1; i <= r; i++)
w[i] += x;
for (int i = in[l] + 1; i < in[r]; i++)
addtag[i] += x;
}
inline int query(int x) {
return w[x] + addtag[in[x]];
}
int main() {
n = read();
cnt = sqrt(n);
for (int i = 1; i <= n; i++)
w[i] = read();
for (int i = 1; i <= n; i++)
in[i] = (i - 1) / cnt + 1;
for (int i = 1; i <= n; i++) {
opt = read();
a = read();
b = read();
c = read();
if (opt == 0)
add(a, b, c);
else
printf("%d\n", query(b));
}
}
//1.l和r在同一个块内，那么只需要暴力的修改l到r便可
//2.l和r不在同一个块，那么要分别对l和r所在的块进行修改
//3.这一种是第2种的一个分支情况，l和r之间还有很多个块，因为这些块都是被整体修改，所以直接打到标记
分块2
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 5e4 + 3;
vector<int> blo[505];
int w[maxn], in[maxn], addtag[maxn];
int n, opt, a, b, c, cnt;
inline ll read() {
ll x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch <= '9' && ch >= '0') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
inline void resort(int x) {
blo[x].clear();
for (int i = (x - 1) * cnt + 1; i <= x * cnt; i++)
blo[x].push_back(w[i]);
sort(blo[x].begin(), blo[x].end());
}
inline void add(int l, int r, int x) {
for (int i = l; i <= min(in[l]*cnt, r); i++)
w[i] += x;
resort(in[l]);
if (in[l] != in[r])
for (int i = (in[r] - 1) * cnt + 1; i <= r; i++)
w[i] += x;
resort(in[r]);
for (int i = in[l] + 1; i < in[r]; i++)
addtag[i] += x;
}
inline int query(int l, int r, int x) {
int ans1 = 0;
for (int i = l; i <= min(r, in[l]*cnt); i++)
if (w[i] + addtag[in[i]] < x)
ans1++;
if (in[l] != in[r])
for (int i = (in[r] - 1) * cnt + 1; i <= r; i++)
if (w[i] + addtag[in[i]] < x)
ans1++;
for (int i = in[l] + 1; i < in[r]; i++) {
int s = x - addtag[i];
ans1 += lower_bound(blo[i].begin(), blo[i].end(), s) - blo[i].begin();
}
return ans1;
}
int main() {
n = read();
cnt = sqrt(n);
for (int i = 1; i <= n; i++)
w[i] = read();
for (int i = 1; i <= n; i++) {
in[i] = (i - 1) / cnt + 1;
blo[in[i]].push_back(w[i]);
}
for (int i = 1; i <= in[n]; i++)
sort(blo[i].begin(), blo[i].end());
for (int i = 1; i <= n; i++) {
opt = read();
a = read();
b = read();
c = read();
if (opt == 0)
add(a, b, c);
else
printf("%d\n", query(a, b, c * c));
}
return 0;
}
//添加过程同分块1
//二分查找来找到第一个大于给定数值的数的位置，然后就可以算出一个块内有多少数是小于给定值的
分块3
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 1e5 + 3;
set<int> blo[1005];
int n, opt, a, b, c, cnt;
int w[maxn], addtag[1005], in[maxn];
inline ll read() {
ll x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch <= '9' && ch >= '0') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
inline void add(int l, int r, int x) {
for (int i = l; i <= min(r, in[l]*cnt); i++) {
blo[in[l]].erase(w[i]);
w[i] += x;
blo[in[l]].insert(w[i]);
}
if (in[l] != in[r])
for (int i = (in[r] - 1) * cnt + 1; i <= r; i++) {
blo[in[r]].erase(w[i]);
w[i] += x;
blo[in[r]].insert(w[i]);
}
for (int i = in[l] + 1; i < in[r]; i++)
addtag[i] += x;
}
inline int query(int l, int r, int x) {
int ans = -1;
for (int i = l; i <= min(in[l]*cnt, r); i++)
if (w[i] + addtag[in[i]] < x)
ans = max(ans, w[i] + addtag[in[i]]);
if (in[l] != in[r])
for (int i = (in[r] - 1) * cnt + 1; i <= r; i++)
if (w[i] + addtag[in[i]] < x)
ans = max(ans, w[i] + addtag[in[i]]);
for (int i = in[l] + 1; i < in[r]; i++) {
int s = x - addtag[i];
set<int>::iterator it = blo[i].lower_bound(s);
if (it == blo[i].begin())
continue;
--it;
ans = max(ans, *it + addtag[i]);
}
return ans;
}
int main() {
n = read();
cnt = 1000;
for (int i = 1; i <= n; i++)
w[i] = read();
for (int i = 1; i <= n; i++) {
in[i] = (i - 1) / cnt + 1;
blo[in[i]].insert(w[i]);
}
for (int i = 1; i <= n; i++) {
opt = read();
a = read();
b = read();
c = read();
if (opt == 0)
add(a, b, c);
else
printf("%d\n", query(a, b, c));
}
return 0;
}
//添加过程相同
//维护一个有序集合set，前驱查找用迭代器， 二分查找x（因为set是有序的），如果x=l输出-1
分块4
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 5e4 + 3;
int n, cnt, opt, l, r, k;
int in[maxn], arr[maxn], addtag[500], sum[500];
inline ll read() {
ll x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch <= '9' && ch >= '0') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
inline void add(int l, int r, int k) {
for (int i = l; i <= min(r, in[l]*cnt); i++) {
arr[i] += k;
sum[in[i]] += k;
}
if (in[l] != in[r]) {
for (int i = (in[r] - 1) * cnt + 1; i <= r; i++) {
arr[i] += k;
sum[in[i]] += k;
}
}
for (int i = in[l] + 1; i < in[r]; i++)
addtag[i] += k;
}
inline int query(int l, int r, int modd) {
int ans = 0;
for (int i = l; i <= min(r, in[l]*cnt); i++)
ans = arr[i] % modd + addtag[in[i]] % modd + ans % modd;
if (in[l] != in[r])
for (int i = (in[r] - 1) * cnt + 1; i <= r; i++)
ans = arr[i] % modd + addtag[in[i]] % modd + ans % modd;
for (int i = in[l] + 1; i < in[r]; i++)
ans = addtag[i] % modd * cnt % modd + sum[i] % modd + ans % modd;
return ans % modd;
}
int main() {
n = read();
cnt = sqrt(n);
for (int i = 1; i <= n; i++) {
arr[i] = read();
in[i] = (i - 1) / cnt + 1;
sum[in[i]] += arr[i];
}
for (int i = 1; i <= n; i++) {
opt = read();
l = read();
r = read();
k = read();
if (opt == 0)
add(l, r, k);
else
printf("%d\n", query(l, r, k + 1));
}
}
//添加同上
//每一个块内都维护一个sum，表示这个块内所有元素的和
//在进行修改区间两端的两个特殊的块的时候修改一个加一个
//到最后查询的时候sum+addtag*cnt，cnt表示块的大小
分块5
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 5e4 + 3;
int n, cnt, opt, l, r, k;
int in[maxn], arr[maxn], sum[500];
bool flag[500];
inline ll read() {
ll x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch <= '9' && ch >= '0') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
inline void solve_sqrt(int x) {
if (flag[x])
return ;
flag[x] = 1;
sum[x] = 0;
for (int i = (x - 1) * cnt + 1; i <= x * cnt; i++) {
arr[i] = sqrt(arr[i]);
sum[x] += arr[i];
if (arr[i] > 1)
flag[x] = 0;
}
}
inline void update(int l, int r) {
for (int i = l; i <= min(r, in[l]*cnt); i++) {
sum[in[i]] -= arr[i];
arr[i] = sqrt(arr[i]);
sum[in[i]] += arr[i];
}
if (in[l] != in[r]) {
for (int i = (in[r] - 1) * cnt + 1; i <= r; i++) {
sum[in[i]] -= arr[i];
arr[i] = sqrt(arr[i]);
sum[in[i]] += arr[i];
}
}
for (int i = in[l] + 1; i < in[r]; i++)
solve_sqrt(i);
}
inline int query(int l, int r) {
int ans = 0;
for (int i = l; i <= min(in[l]*cnt, r); i++)
ans += arr[i];
if (in[r] != in[l])
for (int i = (in[r] - 1) * cnt + 1; i <= r; i++)
ans += arr[i];
for (int i = in[l] + 1; i < in[r]; i++)
ans += sum[i];
return ans;
}
int main() {
n = read();
cnt = sqrt(n);
for (int i = 1; i <= n; i++) {
arr[i] = read();
in[i] = (i - 1) / cnt + 1;
}
for (int i = 1; i <= n; i++)
sum[in[i]] += arr[i];
for (int i = 1; i <= n; i++) {
opt = read();
l = read();
r = read();
k = read();
if (opt == 0)
update(l, r);
else
printf("%d\n", query(l, r));
}
}
//一段区间进行至多5次区间开方就会变成1或者0（上网是个好东西）
//只需要暴力的对区间进行修改
//原因：1和0开方后还是本身，所以对于已经全部变为1或者0的区间就不必再开方了
分块6
#include <bits/stdc++.h>
#define ll long long
#define MAXN 100005
using namespace std;
int n, d, nn, opt, l, r, c;
int a[MAXN << 1];
vector<int> v[1005];
inline ll read() {
ll x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch <= '9' && ch >= '0') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
inline void mer() { //把所有元素还原到a数组中
n = 0;
for (int i = 1; i <= d + 1; i++) {
if (v[i].empty())
break;
for (int j = 0; j < v[i].size(); j++)
a[++n] = v[i][j];
v[i].clear();
}
}
inline void div() { //把a数组元素分配到各个块中
d = sqrt(n);
for (int i = 1; i <= n; i++)
v[(i - 1) / d + 1].push_back(a[i]);
}
inline int Get(int wh) {
for (int i = 1; i <= d + 1; i++) {
if (wh > v[i].size())
wh -= v[i].size();
else
return v[i][wh - 1];
}
}
inline void Ins(int wh, int x) {
for (int i = 1; i <= d + 1; i++) {
if (wh > v[i].size())
wh -= v[i].size();
else {
v[i].insert(v[i].begin() + wh - 1, x); //插入
if (v[i].size() > 10 * d)
mer(), div(); //重排
return ;
}
}
}
int main() {
n = read();
for (int i = 1; i <= n; i++)
a[i] = read();
div();
nn = n;
for (int i = 1; i <= nn; i++) {
opt = read();
l = read();
r = read();
c = read();
if (!opt)
Ins(l, r);
else
cout << Get(r) << endl;
}
return 0;
}
//经典快读~~~
//单点插入使用vector(insert)非常方便，但要注意插入后每个块的长度
//如果多次插入同一块，会导致块过长，时间复杂度过高，失去分块的意义
//所以当一个块的元素量是原来的10倍时，就要重新分块
分块7
#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
const int mod = 10007;
ll n, a[maxn], pos[maxn], m[maxn], sum[maxn], block;
inline ll read() { //经典快读
ll x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch <= '9' && ch >= '0') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
void reset(int x) { //更新不完整块中的每个值
ll p = pos[x];
for (int i = (p - 1) * block + 1; i <= min(p * block, n); i++)
a[i] = (m[p] * a[i] + sum[p]) % mod;
m[p] = 1;
sum[p] = 0; //更新该块的sum,m
}
void update(int l, int r, int c, int opt) { //加和乘
reset(l); //更新左边不完整块
for (int i = l; i <= min(pos[l]*block, (ll)r); i++) { //暴力加或乘
if (opt == 0)
a[i] = (a[i] + c) % mod;
else
a[i] = (a[i] * c) % mod;
}
if (pos[l] != pos[r]) {
reset(r); //更新右边不完整块
for (int i = (pos[r] - 1) * block + 1; i <= r; i++) //暴力加或乘
if (opt == 0)
a[i] = (a[i] + c) % mod;
else
a[i] = (a[i] * c) % mod;
}
for (int i = pos[l] + 1; i <= pos[r] - 1; i++) //完整块处理
if (opt == 0)
sum[i] = (sum[i] + c) % mod; //加
else { //乘
sum[i] = (sum[i] * c) % mod; //sum要乘 关键
m[i] = (m[i] * c) % mod; //系数要乘
}
}
int main() {
n = read();
block = sqrt(n);
for (int i = 1; i <= n; i++) {
a[i] = read();
a[i] %= mod;
pos[i] = (i - 1) / block + 1;
m[pos[i]] = 1; //系数初始化为1
}
int opt, l, r, c;
memset(sum, 0, sizeof(sum));
for (int i = 1; i <= n; i++) {
opt = read();
l = read();
r = read();
c = read();
if (opt == 0 || opt == 1)
update(l, r, c, opt);
else
printf("%lld\n", (a[r]*m[pos[r]] + sum[pos[r]]) % mod);
}
return 0;
}
//加法操作会影响乘法操作的值
//用sum[]存整块的加数，m[]存整块的乘的系数。容易知道，乘的时候sum也要乘
//数值x第一个操作加c1,数值变成了x+c1。用sum[]存c1
//第二个操作乘以c2，数值变成(x+c1)*c2,可以理解为c2*x+c1*c2，用sum[]存c1*c2,m[]存c2
//注意不完整块的处理，每次不完整的块要更新a[]的值，因为可能前面的操作把它当成完整块处理
//不完整块要暴力加，所以要变成最新的值才可以暴力加，只有始终为完整块的数可以不用更新
分块8
#include <bits/stdc++.h>
#define MAXN 100005
#define ll long long
using namespace std;
int n, d;
int a[MAXN], b[MAXN], f[500];
bool v[500];
inline ll read() { //经典快读
ll x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch <= '9' && ch >= '0') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
int FF(int x) { //取出x位置的值
return v[b[x]] ? f[b[x]] : a[x];
}
int fun(int l, int r, int c) { //计数&修改
int ans(0);
for (int i = l; i <= r && i <= n; ++i)
ans += FF(i) == c, a[i] = c;
return ans;
}
int Get(int l, int r, int c) {
int ans(0);
if (b[l] == b[r]) {
if (v[b[l]] && f[b[l]] == c)
return r - l + 1;
if (v[b[l]]) {
fun((b[l] - 1) * d + 1, l - 1, f[b[l]]);
fun(r + 1, b[l] * d, f[b[l]]);
}
ans = fun(l, r, c);
v[b[l]] = 0;
return ans;
}
if (v[b[l]])
fun((b[l] - 1) * d + 1, l - 1, f[b[l]]);
ans += fun(l, b[l] * d, c);
v[b[l]] = 0;
if (v[b[r]])
fun(r + 1, min(b[r] * d, n), f[b[r]]);
ans += fun((b[r] - 1) * d + 1, r, c);
v[b[r]] = 0;
for (int i = b[l] + 1; i <= b[r] - 1; ++i) {
if (!v[i]) {
for (int j = (i - 1) * d + 1; b[j] == i; ++j) {
ans += a[j] == c;
a[j] = c;
}
v[i] = 1;
f[i] = c;
} else {
if (f[i] == c)
ans += d;
else
f[i] = c;
}
}
return ans;
}
int main() {
n = read();
d = (int)sqrt(n);
for (int i = 1; i <= n; ++i) {
a[i] = read();
b[i] = (i - 1) / d + 1;
}
for (int i = 1; i <= n; ++i) {
int l, r, c;
l = read();
r = read();
c = read();
printf("%d\n", Get(l, r, c));
}
return 0;
}
//查询+修改不过多赘述
分块9
#include <bits/stdc++.h>
#define F(i,a,b) for(int i=a;i<=b;i++)
#define D(i,a,b) for(int i=a;i>=b;i--)
#define ms(i,a) memset(a,i,sizeof(a))
#define st(x) ( (x-1)*B+1)
#define ed(x) ( min(n, x*B))
#define bl(x) ( (x-1)/B+1)
#define sum(x) ( (n-1)/B+1)
using namespace std;
int inline read() {
int x = 0 ;
char c = getchar();
while (c < '0' || c > '9')
c = getchar();
while (c >= '0' && c <= '9')
x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
return x;
}
int const maxn = 100003;
int n, m, B, cnt[maxn][401], v[401][401], num[401][401];
int a[maxn], b[maxn], c[maxn], tmp[maxn];
int query(int l, int r) {
int x = bl(l);
int y = bl(r);
int V, NUM = 0;
if (x + 1 <= y - 1) {
V = v[x + 1][y - 1];
NUM = num[x + 1][y - 1];
}
int s, e;
tmp[0] = 0;
if (x == y) {
F(i, l, r) {
tmp[++tmp[0]] = a[i];
c[a[i]]++;
}
} else {
s = l;
e = ed(x);
F(i, s, e) {
tmp[++tmp[0]] = a[i];
c[a[i]]++;
}
s = st(y);
e = r;
F(i, s, e) {
tmp[++tmp[0]] = a[i];
c[a[i]]++;
}
}
F(i, 1, tmp[0]) {
int t = x + 1 <= y - 1 ? cnt[tmp[i]][y - 1] - cnt[tmp[i]][x] : 0;
if (c[tmp[i]] + t > NUM || (c[tmp[i]] + t == NUM && V > tmp[i])) {
V = tmp[i];
NUM = c[tmp[i]] + t;
}
}
F(i, 1, tmp[0])
c[tmp[i]] = 0;
return V;
}
int main() {
n = read();
m = n;
F(i, 1, n)
a[i] = b[i] = read();
sort(b + 1, b + n + 1);
int k = unique(b + 1, b + n + 1) - b - 1;
F(i, 1, n)
a[i] = lower_bound(b + 1, b + k + 1, a[i]) - b;
B = (int)sqrt(n);
int t = sum(n);
F(i, 1, t) {
ms(0, c);
int s = st(i);
int e = ed(i);
F(j, s, e)
c[a[j]]++;
F(j, 1, k)
cnt[j][i] = cnt[j][i - 1] + c[j];
}
F(i, 1, t) {
ms(0, c);
F(j, i, t) {
int V = v[i][j - 1];
int NUM = num[i][j - 1];
int s = st(j);
int e = ed(j);
F(p, s, e) {
c[a[p]]++ ;
if (c[a[p]] > NUM || (NUM == c[a[p]] && a[p] < V)) {
V = a[p];
NUM = c[a[p]];
}
}
v[i][j] = V;
num[i][j] = NUM;
}
}
int x = 0;
ms(0, c);
while (m--) {
int l = read();
int r = read();
x = 1;
if (l > r)
swap(l, r);
printf("%d\n", x = b[query(l, r)]);
}
return 0;
}
//纯查询众数，也就花了1天
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