数列分块入门 (1 ~ 7)

分块

6277. 数列分块入门 1

分块思想

我们把每m个元素分成一块，所以我们总共的块数就是n/mn / mn/m块，一般情况下我们取m=nm = \sqrt{n}m=n.对于区间加操作，我们可以先暴力左右两边，然后对于中间的整块的部分的加减，我们累加在块的标记上，然后我们每次查询的时候只要，每个元素的值加上这个块的标记值，就可以得到我们的答案了。

复杂度分析

每次操作我们修改的最多的元素最多就是O(n)O(\sqrt {n})O(n)级别的个数，时间复杂读也就是O(n)O(\sqrt n)O(n)级别的，查询的时间复杂度是O(1)O(1)O(1)的，最多有nnn个操作，整体上是O(nn)O(n \sqrt{n})O(nn)的，于是一个暴力而优美的分块算法就出现了，简单的思想，\sqrt { }级别的优化。

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define endl '\n'using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}void print(ll x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}const int N = 5e4 + 10;int value[N], bl[N], tag[N], block, n;void add(int l, int r, int c) {for(int i = l; i <= bl[l] * block && i <= r; i++) {//对前面的部分进行暴力修改。value[i] += c;}if(bl[l] != bl[r]) {//如果这两个块不在同一个分块中才要进行后面的暴力修改，否则将会重复累加。for(int i = block * (bl[r] - 1) + 1; i <= r; i++) {value[i] += c;}}for(int i = bl[l] + 1; i <= bl[r] - 1; i++) {//对每个块进行区间修改。tag[i] += c;}//严格来说每次修改的复杂度最多将会是3 * sqrt(n)
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read(); block = sqrt(n);//block是分块的大小for(int i = 1; i <= n; i++) {value[i] = read();bl[i] = (i - 1) / block + 1;//每个位置在分块中的位置。}for(int i = 1; i <= n; i++) {int op = read(), l = read(), r = read(), c = read();if(!op) {add(l, r, c);}else {printf("%d\n", value[r] + tag[bl[r]]);//他当前的值加上分块的累加值。}}return 0;
}

6278. 数列分块入门 2

想法

查询操作：对于每一个块我们开一个数组来维护其有序的状态，所以对于每一个整块我们可以通过二分去得到我们的答案，对于两头的块，我们可以暴力去得到满足条件的数。

修改操作：同样的对于整块的我们还是在一个区间标记加上其修改值，对于两头的操作我们需要额外单独的操作，记得操作完之后要把原来的存区间有序的数组重新排序，得到一个新的区间有序数组。

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define endl '\n'using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}void print(ll x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}const int N = 5e4 + 10;int value[N], bl[N], tag[N], n, block, m;vector<int> elem[N];void reset(int x) {//最好还是写一个函数，一开始我就是没写函数然后wa了一次。elem[x].clear();for(int i = (x - 1) * block + 1; i <= x * block && i <= n; i++)elem[x].pb(value[i]);sort(elem[x].begin(), elem[x].end());
}void add(int l, int r, int x) {for(int i = l; i <= r && i <= bl[l] * block; i++)value[i] += x;reset(bl[l]);if(bl[l] != bl[r]) {for(int i = (bl[r] - 1) * block + 1; i <= r; i++)value[i] += x;reset(bl[r]);}for(int i = bl[l] + 1; i <= bl[r] - 1; i++)tag[i] += x;
}int query(int l, int r, int x) {int sum = 0;for(int i = l; i <= r && i <= bl[l] * block; i++)if(value[i] + tag[bl[i]] < x)sum++;if(bl[l] != bl[r])for(int i = (bl[r] - 1) * block + 1; i <= r; i++)if(value[i] + tag[bl[i]] < x)sum++;for(int i = bl[l] + 1; i <= bl[r] - 1; i++) {sum += lower_bound(elem[i].begin(), elem[i].end(), x - tag[i]) - elem[i].begin();}return sum;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read(); block = sqrt(n);for(int i = 1; i <= n; i++) {value[i] = read();bl[i] = (i + block - 1) / block;elem[bl[i]].pb(value[i]);}m = bl[n];for(int i = 1; i <= m; i++) sort(elem[i].begin(), elem[i].end());for(int i = 1; i <= n; i++) {int op = read(), l = read(), r = read(), c = read();if(op & 1) {printf("%d\n", query(l, r, c * c));}else {add(l, r, c);}}return 0;
}

6279. 数列分块入门 3

想法

这题应该是跟上一个类似，就是queryqueryquery操作稍微变化一下。

用上面的代码修改的时候要注意，这题数据变大了，需要改成1e5+101e5 + 101e5+10，不然就跟我一样入坑了。

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define endl '\n'using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}void print(ll x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}const int N = 1e5 + 10;int value[N], bl[N], tag[N], n, block, m;vector<int> elem[N];void reset(int x) {elem[x].clear();for(int i = (x - 1) * block + 1; i <= x * block && i <= n; i++)elem[x].pb(value[i]);sort(elem[x].begin(), elem[x].end());
}void add(int l, int r, int x) {for(int i = l; i <= r && i <= bl[l] * block; i++)value[i] += x;reset(bl[l]);if(bl[l] != bl[r]) {for(int i = (bl[r] - 1) * block + 1; i <= r; i++)value[i] += x;reset(bl[r]);}for(int i = bl[l] + 1; i <= bl[r] - 1; i++)tag[i] += x;
}ll query(int l, int r, int x) {ll ans = -1e10;for(int i = l; i <= r && i <= bl[l] * block; i++)if(value[i] + tag[bl[i]] < x && value[i] + tag[bl[i]] > ans)ans = value[i] + tag[bl[i]];if(bl[l] != bl[r])for(int i = (bl[r] - 1) * block + 1; i <= r; i++)if(value[i] + tag[bl[i]] < x && value[i] + tag[bl[i]] > ans)ans = value[i] + tag[bl[i]];for(int i = bl[l] + 1; i <= bl[r] - 1; i++) {auto p = lower_bound(elem[i].begin(), elem[i].end(), x - tag[i]);if(p == elem[i].begin()) continue;p--;ans = max(ans, 1ll * (*p + tag[i]));}return ans == -1e10 ? -1 : ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read(); block = sqrt(n);for(int i = 1; i <= n; i++) {value[i] = read();bl[i] = (i + block - 1) / block;elem[bl[i]].pb(value[i]);}m = bl[n];for(int i = 1; i <= m; i++) sort(elem[i].begin(), elem[i].end());for(int i = 1; i <= n; i++) {int op = read(), l = read(), r = read(), c = read();if(op & 1) {printf("%d\n", query(l, r, c));}else {add(l, r, c);}}return 0;
}

6280. 数列分块入门 4

想法

仿照题二，我们可以再引入一个数组来代表当前块的总和，之后的操作就变得简单了。

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define endl '\n'using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}void print(ll x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}const int N = 1e5 + 10;ll a[N], sum[N], tag[N];int n, block, bl[N];void update(int l, int r, int x) {for(int i = l; i <= bl[l] * block && i <= r; i++) {a[i] += x;sum[bl[i]] += x;}if(bl[l] != bl[r]) {for(int i = (bl[r] - 1) * block + 1; i <= r; i++) {a[i] += x;sum[bl[i]] += x;}}for(int i = bl[l] + 1; i <= bl[r] - 1; i++)tag[i] += x;
}int query(int l, int r, int mod) {ll ans = 0;for(int i = l; i <= bl[l] * block && i <= r; i++) {ans = (ans + a[i] + tag[bl[i]]) % mod;}if(bl[l] != bl[r]) {for(int i = (bl[r] - 1) * block + 1; i <= r; i++) {ans = (ans + a[i] + tag[bl[i]]) % mod;}}for(int i = bl[l] + 1; i <= bl[r] - 1; i++) {ans = (ans + sum[i] + tag[i] * block) % mod;}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read(); block = sqrt(n);for(int i = 1; i <= n; i++) {a[i] = read();bl[i] = (i + block - 1) / block;sum[bl[i]] += a[i];}for(int i = 1; i <= n; i++) {int op = read(), l = read(), r = read(), c = read();if(op & 1) {printf("%d\n", query(l, r, c + 1));}else {update(l, r, c);}}return 0;
}

6281. 数列分块入门 5

想法

一开始看到这个题目感觉分块无从下手啊，区间开方操作，结果一看他们ac代码，原来是开根号，，，

既然是开根号那就简单了，学过线段树应该写过这类题，我们只要加一个区间tag来标记区间内所有的数是否全部变成0和1即可。然后再加一个区间和，对于tag标记为，区间所有的数都是0 和1的我们可以直接查询区间和，也可以跳过他的开方显然有1=1\sqrt {1} = 11=1， 0=0\sqrt {0} = 00=0

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define endl '\n'using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}void print(ll x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}const int N = 1e5 + 10;int a[N], sum[N], tag[N];int n, block, bl[N];void update(int l, int r) {for(int i = l; i <= r && i <= bl[l] * block; i++) {if(a[i] == 0 || a[i] == 1) continue;sum[bl[i]] -= a[i];a[i] = sqrt(a[i]);sum[bl[i]] += a[i];if(a[i] == 1 || a[i] == 0) tag[bl[i]]++;}if(bl[l] != bl[r]) {for(int i = (bl[r] - 1) * block + 1; i <= r; i++) {if(a[i] == 0 || a[i] == 1) continue;sum[bl[i]] -= a[i];a[i] = sqrt(a[i]);sum[bl[i]] += a[i];if(a[i] == 1 || a[i] == 0) tag[bl[i]]++;}}for(int i = bl[l] + 1; i <= bl[r] - 1; i++) {if(tag[i] == block) continue;else {for(int j = (i - 1) * block + 1; j <= i * block; j++) {if(a[j] == 0 || a[j] == 1) continue;sum[bl[j]] -= a[j];a[j] = sqrt(a[j]);sum[bl[j]] += a[j];if(a[i] == 1 || a[i] == 0) tag[bl[i]]++;}}}
}int query(int l, int r) {int ans = 0;for(int i = l; i <= r && i <= bl[l] * block; i++)ans += a[i];if(bl[l] != bl[r]) {for(int i = (bl[r] - 1) * block + 1; i <= r; i++)ans += a[i];}for(int i = bl[l] + 1; i <= bl[r] - 1; i++) {if(tag[i] == block) ans += sum[i];else {for(int j = (i - 1) * block + 1; j <= i * block; j++)ans += a[j];}}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read(); block = sqrt(n);for(int i = 1; i <= n; i++) {a[i] = read();bl[i] = (i + block - 1) / block;sum[bl[i]] += a[i];if(a[i] == 0 || a[i] == 1) tag[bl[i]]++;}for(int i = 1; i <= n; i++) {int op = read(), l = read(), r = read(), c = read();if(op & 1) {printf("%d\n", query(l, r));}else {update(l, r);}}return 0;
}

6282. 数列分块入门 6

想法

这里引入了一个暴力的rebuildrebuildrebuild，不愧是分块，依旧是如此的暴力。

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define endl '\n'using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}void print(ll x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}const int N = 2e5 + 10;int value[N], n, m, block;vector<int> a[N];pii query(int x) {int now = 1;while(x > a[now].size()) {x -= a[now].size(), now++;}return mp(now, x - 1);
}void rebuild() {int num = 0;for(int i = 1; i <= m; i++) {for(int j = 0; j < a[i].size(); j++) {value[++num] = a[i][j];}a[i].clear();//注意一定要clear，我就在这里找bug找了半小时。}block = sqrt(num), m = (num + block - 1) / block;for(int i = 1; i <= num; i++) {a[(i + block - 1) / block].pb(value[i]);}
}void insert(int pos, int value) {pii p = query(pos);a[p.first].insert(a[p.first].begin() + p.second, value);if(a[p.first].size() > 20 * block)rebuild();
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read();block = sqrt(n), m = (n + block - 1) / block;for(int i = 1; i <= n; i++) {value[i] = read();a[(i + block - 1) / block].pb(value[i]);}for(int i = 1; i <= n; i++) {int op = read(), l = read(), r = read(), c = read();if(op & 1) {pii ans = query(r);printf("%d\n", a[ans.first][ans.second]);}else {insert(l, r);}}return 0;
}

6283. 数列分块入门 7

想法

暴力两头，区间操作中间。

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define endl '\n'using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}void print(ll x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}const int N = 1e5 + 10, mod = 1e4 + 7;int a[N], add[N], mult[N], bl[N], n, block;void init(int bl) {for(int i = (bl - 1) * block + 1; i <= bl * block; i++) {a[i] = (a[i] * mult[bl]) % mod;a[i] = (a[i] + add[bl]) % mod;}mult[bl] = 1, add[bl] = 0;
}void Add(int l, int r, int x) {init(bl[l]);for(int i = l; i <= r && i <= bl[l] * block; i++)a[i] = (a[i] + x) % mod;if(bl[l] != bl[r]) {init(bl[r]);for(int i = (bl[r] - 1) * block + 1; i <= r; i++)a[i] = (a[i] + x) % mod;}for(int i = bl[l] + 1; i <= bl[r] - 1; i++)add[i] = (add[i] + x) % mod;
}void Mult(int l, int r, int x) {init(bl[l]);for(int i = l; i <= r && i <= bl[l] * block; i++)a[i] = (a[i] * x) % mod;if(bl[l] != bl[r]) {init(bl[r]);for(int i = (bl[r] - 1) * block + 1; i <= r; i++)a[i] = (a[i] * x) % mod;}for(int i = bl[l] + 1; i <= bl[r] - 1; i++)mult[i] = (mult[i] * x) % mod, add[i] = (add[i] * x) % mod;
}int query(int pos) {return (((a[pos] * mult[bl[pos]])) % mod + add[bl[pos]]) % mod;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read();block = sqrt(n);for(int i = 1; i <= n; i++) {a[i] = read() % mod;bl[i] = (i + block - 1) / block;mult[i] = 1;}for(int i = 1; i <= n; i++) {int op = read(), l = read(), r = read(), x = read();if(!op) {Add(l, r, x);} else if(op & 1) {Mult(l, r, x);} else {printf("%d\n", query(r));}}return 0;
}