[Loj] 数列分块入门 1

数列分块入门 1

区间加 + 单点查询

#include <iostream>

#include <cstdio>

#include <cmath>

using namespace std;

const int N = 5e4 + ;

#define gc getchar()

inline int read() {

    int x = ; char c = gc;

    while(c < '' || c > '') c = gc;

    while(c >= '' && c <= '') x = x *  + c - '', c = gc;

    return x;

}

int A[N], Add[N], bel[N];

int n, block, cnt;

void Sec_G(int x, int y, int w) {

    if(bel[x] == bel[y]) for(int i = x; i <= y; i ++) A[i] += w;

    else {

        for(int i = x; i <= bel[x] * block; i ++) A[i] += w;

        for(int i = (bel[y] - ) * block + ; i <= y; i ++) A[i] += w;

    }

    for(int i = bel[x] + ; i < bel[y]; i ++) Add[i] += w;

}

int main() {

    n = read();

    block = sqrt(n);

    for(int i = ; i <= n; i ++) A[i] = read();

    for(int i = ; i <= n; i ++) bel[i] = (i - ) / block + ;

    if(n % block) cnt = n / block + ;

    else cnt = n / block;

    int T = n;

    while(T --) {

        int opt = read(), l = read(), r = read(), c = read();

        if(!opt) Sec_G(l, r, c);

        else cout << A[r] + Add[bel[r]] << endl;

    }

    return ;

}

数列分块入门 2

https://loj.ac/problem/6278

区间加法，询问区间内小于某个值 x 的元素个数

用B[]记录A[], B[]数组中为排好序的A[]的映射

那么每次可以对每一块进行二分查找

#include <iostream>

#include <cstdio>

#include <cmath>

#include <algorithm> 

using namespace std;

const int N = 5e4 + ;

#define gc getchar()

inline int read() {

    int x = ; char c = gc;

    while(c < '' || c > '') c = gc;

    while(c >= '' && c <= '') x = x *  + c - '', c = gc;

    return x;

}

int A[N], B[N], Add[N], bel[N];

int n, block, cnt;

void Work_sort(int x) {

    int l = (x - ) * block + , r = min(l + block - , n);

    for(int i = l; i <= r; i ++) B[i] = A[i];

    sort(B + l, B + r + );

}

void Sec_G(int x, int y, int w) {

    if(bel[x] == bel[y]) {

        for(int i = x; i <= y; i ++) A[i] += w;

        Work_sort(bel[x]);

    }

    else {

        for(int i = x; i <= bel[x] * block; i ++) A[i] += w; Work_sort(bel[x]);

        for(int i = (bel[y] - ) * block + ; i <= y; i ++) A[i] += w; Work_sort(bel[y]);

    }

    for(int i = bel[x] + ; i < bel[y]; i ++) Add[i] += w;

}

inline int Calc(int x, int w) {

    int l = (x - ) * block + , r = min(l + block - , n), ret = ;

    while(l <= r) {

        int mid = (l + r) >> ;

        if(B[mid] + Add[x] < w) ret = mid, l = mid + ;

        else r = mid - ;

    }

    return ret ? (ret - (x - ) * block) : ;

}

inline int Sec_A(int x, int y, int w) {

    int ret();

    if(bel[x] == bel[y]) {

        for(int i = x; i <= y; i ++) if(A[i] + Add[bel[x]] < w) ret ++;

        return ret;

    } else {

        for(int i = x; i <= bel[x] * block; i ++) if(A[i] + Add[bel[x]] < w) ret ++;

        for(int i = (bel[y] - ) * block + ; i <= y; i ++)

        if(A[i] + Add[bel[y]] < w)

        ret ++;

    }

    for(int i = bel[x] + ; i < bel[y]; i ++)

    ret += Calc(i, w);

    return ret;

}

int main() {

    n = read();

    block = sqrt(n);

    for(int i = ; i <= n; i ++) A[i] = read();

    for(int i = ; i <= n; i ++) bel[i] = (i - ) / block + ;

    if(n % block) cnt = n / block + ;

    else cnt = n / block;

    for(int i = ; i <= cnt; i ++) Work_sort(i);

    int T = n;

    while(T --) {

        int opt = read(), l = read(), r = read(), c = read();

        if(!opt) Sec_G(l, r, c);

        else cout << Sec_A(l, r, c * c) << "\n";

    }

    return ;

}

数列分块入门 3

https://loj.ac/problem/6279

区间加法，询问区间内小于某个值 x 的前驱

与2类似，二分查找

#include <iostream>

#include <cstdio>

#include <cmath>

#include <algorithm> 

using namespace std;

const int N = 1e5 + ;

const int oo = ;

#define gc getchar()

inline int read() {

    int x = ; char c = gc;

    while(c < '' || c > '') c = gc;

    while(c >= '' && c <= '') x = x *  + c - '', c = gc;

    return x;

}

int A[N], B[N] = {-}, Add[N], bel[N];

int n, block, cnt;

void Work_sort(int x) {

    int l = (x - ) * block + , r = min(l + block - , n);

    for(int i = l; i <= r; i ++) B[i] = A[i];

    sort(B + l, B + r + );

}

void Sec_G(int x, int y, int w) {

    if(bel[x] == bel[y]) {

        for(int i = x; i <= y; i ++) A[i] += w;

        Work_sort(bel[x]);

    }

    else {

        for(int i = x; i <= bel[x] * block; i ++) A[i] += w; Work_sort(bel[x]);

        for(int i = (bel[y] - ) * block + ; i <= y; i ++) A[i] += w; Work_sort(bel[y]);

    }

    for(int i = bel[x] + ; i < bel[y]; i ++) Add[i] += w;

}

inline int Calc(int x, int w) {

    int l = (x - ) * block + , r = min(l + block - , n), ret = ;

    while(l <= r) {

        int mid = (l + r) >> ;

        if(B[mid] + Add[x] < w) ret = mid, l = mid + ;

        else r = mid - ;

    }

    return B[ret] + Add[x];

}

inline int Sec_A(int x, int y, int w) {

    int ret = -;

    if(bel[x] == bel[y]) {

        for(int i = x; i <= y; i ++) if(A[i] + Add[bel[x]] < w && A[i] + Add[bel[x]] > ret) ret = A[i] + Add[bel[x]];

        return ret;

    } else {

        for(int i = x; i <= bel[x] * block; i ++)

        if(A[i] + Add[bel[x]] < w && A[i] + Add[bel[x]] > ret) ret = A[i] + Add[bel[x]];

        for(int i = (bel[y] - ) * block + ; i <= y; i ++)

        if(A[i] + Add[bel[y]] < w && A[i] + Add[bel[y]] > ret) ret = A[i] + Add[bel[y]];

    }

    for(int i = bel[x] + ; i < bel[y]; i ++) {

        int imp = Calc(i, w);

        if(imp < w && imp > ret) ret = imp;

    }

    return ret;

}

int main() {

    n = read();

    block = sqrt(n);

    for(int i = ; i <= n; i ++) A[i] = read();

    for(int i = ; i <= n; i ++) bel[i] = (i - ) / block + ;

    if(n % block) cnt = n / block + ;

    else cnt = n / block;

    for(int i = ; i <= cnt; i ++) Work_sort(i);

    int T = n;

    while(T --) {

        int opt = read(), l = read(), r = read(), c = read();

        if(!opt) Sec_G(l, r, c);

        else cout << Sec_A(l, r, c) << "\n";

    }

    return ;

}

数列分块入门 4

区间加法，区间求和

https://loj.ac/problem/6280

没什么好说的

#include <iostream>

#include <cstdio>

#include <cmath>

using namespace std;

const int N = 5e4 + ;

#define gc getchar()

inline int read() {

    int x = ; char c = gc;

    while(c < '' || c > '') c = gc;

    while(c >= '' && c <= '') x = x *  + c - '', c = gc;

    return x;

}

#define LL long long

LL A[N], Add[N], bel[N], W[N];

int n, block, cnt, Mod;

void Sec_G(int x, int y, int w) {

    if(bel[x] == bel[y]) for(int i = x; i <= y; i ++) A[i] += w, W[bel[x]] += w;

    else {

        for(int i = x; i <= bel[x] * block; i ++) A[i] += w, W[bel[x]] += w;

        for(int i = (bel[y] - ) * block + ; i <= y; i ++) A[i] += w, W[bel[y]] += w;

    }

    for(int i = bel[x] + ; i < bel[y]; i ++) Add[i] += w, W[i] += w * block;

}

inline int Sec_A(int x, int y) {

    LL ret = ;

    if(bel[x] == bel[y])

        for(int i = x; i <= y; i ++)

            ret += (A[i] + Add[bel[x]]) % Mod;

    else {

        for(int i = x; i <= bel[x] * block; i ++) ret += (A[i] + Add[bel[x]]) % Mod;

        for(int i = (bel[y] - ) * block + ; i <= y; i ++) ret += (A[i] + Add[bel[y]]) % Mod;

    }

    for(int i = bel[x] + ; i < bel[y]; i ++) ret += W[i] % Mod;

    return ret% Mod;

}

int main() {

    n = read();

    block = sqrt(n);

    for(int i = ; i <= n; i ++) A[i] = read();

    for(int i = ; i <= n; i ++) bel[i] = (i - ) / block + , W[bel[i]] += A[i];

    if(n % block) cnt = n / block + ;

    else cnt = n / block;

    int T = n;

    while(T --) {

        int opt = read(), l = read(), r = read(), c = read(); Mod = c + ;

        if(!opt) Sec_G(l, r, c);

        else cout << Sec_A(l, r) << endl;

    }

    return ;

}

数列分块入门 5

区间开方，区间求和

一个数(合理)开几次根后就是0/1了

因此，只需记录每块的最大值，如果最大值是0/1就没必要开根

#include <iostream>

#include <cstdio>

#include <cmath>

using namespace std;

const int N = 5e4 + ;

#define gc getchar()

inline int read() {

    int x = ; char c = gc;

    while(c < '' || c > '') c = gc;

    while(c >= '' && c <= '') x = x *  + c - '', c = gc;

    return x;

}

#define LL long long

LL A[N], Add[N], bel[N], W[N], Max[N];

int n, block, cnt, Mod;

inline void Sec_G(int x, int y) {

    if(bel[x] == bel[y]) {

        if(!Max[bel[x]]) return ;

        if(Max[bel[x]] == ) return ;

        for(int i = x; i <= y; i ++) {

            int C = A[i] - (int) sqrt(A[i]);

            W[bel[x]] -= C; A[i] = (int) sqrt(A[i]);

        }

        LL Max_A = ;

        for(int i = (bel[x] - ) * block + ; i <= bel[x] * block; i ++) Max_A = max(Max_A, A[i]);

        Max[bel[x]] = Max_A;

        return ;

    }

    else {

        if(Max[bel[x]] && Max[bel[x]] != ) {

            for(int i = x; i <= bel[x] * block; i ++) {

                int C = A[i] - (int) sqrt(A[i]);

                W[bel[x]] -= C;

                A[i] = (int) sqrt(A[i]);

            }

            LL Max_A = ;

            for(int i = (bel[x] - ) * block + ; i <= bel[x] * block; i ++) Max_A = max(Max_A, A[i]);

            Max[bel[x]] = Max_A;

        }

        if(Max[bel[y]] && Max[bel[y]] != ) {

            for(int i = (bel[y] - ) * block + ; i <= y; i ++) {

                int C = A[i] - (int) sqrt(A[i]);

                W[bel[y]] -= C;

                A[i] = (int) sqrt(A[i]);

            }

            LL Max_A = ;

            for(int i = (bel[y] - ) * block + ; i <= bel[y] * block; i ++) Max_A = max(Max_A, A[i]);

            Max[bel[y]] = Max_A;

        }

    }

    for(int i = bel[x] + ; i < bel[y]; i ++) {

        if(!Max[i] || Max[i] == ) continue ;

        LL Max_A = ;

        for(int j = (i - ) * block + ; j <= i * block; j ++) {

            int C = A[j] - (int) sqrt(A[j]);

            W[i] -= C;

            A[j] = (int) sqrt(A[j]);

            Max_A = max(Max_A, A[j]);

        }

        Max[i] = Max_A;

    }

}

inline int Sec_A(int x, int y) {

    LL ret = ;

    if(bel[x] == bel[y] && Max[bel[x]]) for(int i = x; i <= y; i ++) ret += A[i];

    else {

        for(int i = x; i <= bel[x] * block && Max[bel[x]]; i ++) ret += A[i];

        for(int i = (bel[y] - ) * block + ; i <= y && Max[bel[y]]; i ++) ret += A[i];

    }

    for(int i = bel[x] + ; i < bel[y]; i ++) ret += W[i];

    return ret;

}

int main() {

    n = read();

    block = sqrt(n);

    for(int i = ; i <= n; i ++) A[i] = read();

    for(int i = ; i <= n; i ++) bel[i] = (i - ) / block + , W[bel[i]] += A[i], Max[bel[i]] = max(Max[bel[i]], A[i]);

    int T = n;

    while(T --) {

        int opt = read(), l = read(), r = read(), c = read();

        if(!opt) Sec_G(l, r);

        else cout << Sec_A(l, r) << "\n";

    }

    return ;

}

数列分块入门 6

https://loj.ac/problem/6282

单点插入，单点询问

数据随机，分块，对于每一块开动态数组，插入 + 查询比较容易实现

如果数据不随机，就有可能加到同一块中的数较多，影响效率

这样可以进行一定的插入操作之后重新分块

#include <iostream>

#include <cstdio>

#include <cmath>

#include <vector>

using namespace std;

const int N = 1e5 + ;

int A[N << ], n;

vector <int> Vec[];

int block, bel[N];

#define gc getchar()

inline int read() {

    int x = ; char c = gc;

    while(c < '' || c > '') c = gc;

    while(c >= '' && c <= '') x = x *  + c - '', c = gc;

    return x;

}

inline void Ins(int x, int a) {

    int now_size(), Whi;

    for(int i = ; ; i ++) {

        int Size = Vec[i].size();

        now_size += Size;

        if(now_size >= x) {

            Whi = i;

            x -= (now_size - Size);

            break;

        }

    }

    Vec[Whi].insert(Vec[Whi].begin() + x - , a);

}

inline int Poi_A(int x) {

    int Whi_, now_size();

    for(int i = ; ; i ++) {

        int Size = Vec[i].size();

        now_size += Size;

        if(now_size >= x) {

            int iii = x - (now_size - Size);

            return Vec[i][iii - ];

        }

    }

}

int main() {

    n = read();

    block = sqrt(n);

    for(int i = ; i <= n; i ++) A[i] = read();

    for(int i = ; i <= n; i ++) bel[i] = (i - ) / block + ;

    for(int i = ; i <= n; i ++) Vec[bel[i]].push_back(A[i]);

    int T = n;

    while(T --) {

        int opt = read(), l = read(), r = read(), c = read();

        if(!opt) Ins(l, r);

        else cout << Poi_A(r) << endl;

    }

    return ;

}

数列分块入门 7

https://loj.ac/problem/6283

区间乘法，区间加法，单点询问

先乘后加，乘的时候相应的加法标记也要乘

#include <iostream>

#include <cstdio>

#include <cmath>

#include <vector>

using namespace std;

const int N = 1e5 + ;

const int Mod = 1e4 + ;

#define LL long long

LL A[N], Mul[N], Add[N], bel[N];

int n, cnt, block;

#define gc getchar()

inline int read() {

    int x = , f = ; char c = gc;

    while(c < '' || c > '') {if(c == '-') f = -; c = gc;}

    while(c >= '' && c <= '') x = x *  + c - '', c = gc;

    return x;

}

inline void Sec_Add(int x, int y, int w) {

    if(bel[x] == bel[y]) {

        for(int i = (bel[x] - ) * block + ; i <= bel[x] * block; i ++) A[i] = (A[i] * Mul[bel[x]] + Add[bel[x]]);

        Add[bel[x]] = ; Mul[bel[x]] = ;

        for(int i = x; i <= y; i ++) A[i] += w, A[i] %= Mod;

        return ;

    } else {

        for(int i = (bel[x] - ) * block + ; i <= bel[x] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod;

        Add[bel[x]] = ; Mul[bel[x]] = ;

        for(int i = x; i <= bel[x] * block; i ++) A[i] += w, A[i] %= Mod;

        for(int i = (bel[y] - ) * block + ; i <= bel[y] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod;

        Add[bel[y]] = ; Mul[bel[y]] = ;

        for(int i = (bel[y] - ) * block + ; i <= y; i ++) A[i] += w, A[i] %= Mod;

    }

    for(int i = bel[x] + ; i < bel[y]; i ++) Add[i] += w, Add[i] %= Mod;

}

inline void Sec_Mul(int x, int y, int w) {

    if(bel[x] == bel[y]) {

        for(int i = (bel[x] - ) * block + ; i <= bel[x] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod;

        Add[bel[x]] = ; Mul[bel[x]] = ;

        for(int i = x; i <= y; i ++) A[i] = (A[i] * w) % Mod;

        return ;

    } else {

        for(int i = (bel[x] - ) * block + ; i <= bel[x] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod;

        Add[bel[x]] = ; Mul[bel[x]] = ;

        for(int i = x; i <= bel[x] * block; i ++) A[i] = (A[i] * w) % Mod;

        for(int i = (bel[y] - ) * block + ; i <= bel[y] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod;

        Add[bel[y]] = ; Mul[bel[y]] = ;

        for(int i = (bel[y] - ) * block + ; i <= y; i ++) A[i] = (A[i] * w) % Mod;

    }

    for(int i = bel[x] + ; i < bel[y]; i ++) Add[i] = (Add[i] * w) % Mod, Mul[i] = (Mul[i] * w) % Mod;

}

int main() {

    n = read();

    for(int i = ; i <= n; i ++) A[i] = read();

    block = sqrt(n);

    for(int i = ; i <= n; i ++) bel[i] = (i - ) / block + , Mul[i] = ;

    int T = n;

    while(T --) {

        int opt = read(), l = read(), r = read(), c = read();

        if(opt == ) Sec_Add(l, r, c);

        else if(opt == ) Sec_Mul(l, r, c);

        else cout << (A[r] * Mul[bel[r]] + Add[bel[r]]) % Mod << "\n";

    }

    return ;

}

数列分块入门 8

https://loj.ac/problem/6284

暴力

区间修改没有什么难度，这题难在区间查询比较奇怪，因为权值种类比较多，似乎没有什么好的维护方法。

模拟一些数据可以发现，询问后一整段都会被修改，几次询问后数列可能只剩下几段不同的区间了。

我们思考这样一个暴力，还是分块，维护每个分块是否只有一种权值，区间操作的时候，对于同权值的一个块就O(1)统计答案，否则暴力统计答案，并修改标记，不完整的块也暴力。

这样看似最差情况每次都会耗费O(n)的时间，但其实可以这样分析：

假设初始序列都是同一个值，那么查询是O(√n)，如果这时进行一个区间操作，它最多破坏首尾2个块的标记，所以只能使后面的询问至多多2个块的暴力时间，所以均摊每次操作复杂度还是O(√n)。

换句话说，要想让一个操作耗费O(n)的时间，要先花费√n个操作对数列进行修改。

初始序列不同值，经过类似分析后，就可以放心的暴力啦。

#include <iostream>

#include <cstdio>

#include <cmath>

#include <cstdlib>

#include <cstring>

using namespace std;

const int N = 1e5 + ;

int bel[N], A[N], bec[N];

int n;

int block;

#define gc getchar()

inline int read() {

    int x = ; char c = gc;

    while(c < '' || c > '') c = gc;

    while(c >= '' && c <= '') x = x *  + c - '', c = gc;

    return x;

}

int Sec_A(int x, int y, int c) {

    int ret();

    if(bel[x] == bel[y]) {

        if(bec[bel[x]] == c) ret = y - x + ;

        else if(bec[bel[x]] == -) for(int i = x; i <= y; i ++) if(A[i] == c) ret ++;

        if(~ bec[bel[x]]) for(int i = (bel[x] - ) * block + ; i < x; i ++) A[i] = bec[bel[x]];

        if(~ bec[bel[x]]) for(int i = y + ; i <= bel[x] * block; i ++) A[i] = bec[bel[x]];

        for(int i = x; i <= y; i ++) A[i] = c;

        bec[bel[x]] = -;

    } else {

        if(bec[bel[x]] == c) ret += bel[x] * block - x + ;

        else if(bec[bel[x]] == -) for(int i = x; i <= bel[x] * block; i ++) if(A[i] == c) ret ++;

        if(bec[bel[y]] == c) ret += y - ((bel[y] - ) * block);

        else if(bec[bel[y]] == -) for(int i = (bel[y] - ) * block + ; i <= y; i ++) if(A[i] == c) ret ++;

        for(int i = bel[x] + ; i < bel[y]; i ++) {

            if(bec[i] == c) ret += block;

            else if(bec[i] == -)

                for(int j = (i - ) * block + ; j <= i * block; j ++)

                    if(A[j] == c) ret ++;

        }

        if(~ bec[bel[x]]) for(int i = (bel[x] - ) * block + ; i < x; i ++) A[i] = bec[bel[i]];

        for(int i = x; i <= bel[x] * block; i ++) A[i] = c;

        bec[bel[x]] = -;

        for(int i = (bel[y] - ) * block + ; i <= y; i ++) A[i] = c;

        if(~ bec[bel[y]]) for(int i = y + ; i <= bel[y] * block; i ++) A[i] = bec[bel[i]];

        bec[bel[y]] = -;

        for(int i = bel[x] + ; i < bel[y]; i ++) bec[i] = c;

    }

    return ret;

}

int main() {

    n = read();

    for(int i = ; i <= n; i ++) bec[i] = -;

    for(int i = ; i <= n; i ++) A[i] = read();

    block = sqrt(n);

    for(int i = ; i <= n; i ++) bel[i] = (i - ) / block + ;

    int T = n;

    while(T --) {

        int l = read(), r = read(), c = read();

        cout << Sec_A(l, r, c) << "\n";

    }

    return ;

}

数列分块入门 9

https://loj.ac/problem/6285

区间众数查询

陈立杰区间众数解题报告

#include<map>

#include<set>

#include<cmath>

#include<stack>

#include<queue>

#include<cstdio>

#include<vector>

#include<cstring>

#include<cstdlib>

#include<iostream>

#include<algorithm>

#define mod 10007

#define pi acos(-1)

#define inf 0x7fffffff

#define ll long long

using namespace std;

ll read() {

    ll x=,f=;

    char ch=getchar();

    while(ch<''||ch>'') {

        if(ch=='-')f=-;

        ch=getchar();

    }

    while(ch>=''&&ch<='') {

        x=x*+ch-'';

        ch=getchar();

    }

    return x*f;

}

int n,blo,id;

int v[],bl[];

int f[][];

map<int,int>mp;

int val[],cnt[];

vector<int>ve[];

void pre(int x) {

    memset(cnt,,sizeof(cnt));

    int mx=,ans=;

    for(int i=(x-)*blo+; i<=n; i++) {

        cnt[v[i]]++;

        int t=bl[i];

        if(cnt[v[i]]>mx||(cnt[v[i]]==mx&&val[v[i]]<val[ans]))

            ans=v[i],mx=cnt[v[i]];

        f[x][t]=ans;

    }

}

int query(int l,int r,int x) {

    int t=upper_bound(ve[x].begin(),ve[x].end(),r)-lower_bound(ve[x].begin(),ve[x].end(),l);

    return t;

}

int query(int a,int b) {

    int ans,mx;

    ans=f[bl[a]+][bl[b]-];

    mx=query(a,b,ans);

    for(int i=a; i<=min(bl[a]*blo,b); i++) {

        int t=query(a,b,v[i]);

        if(t>mx||(t==mx&&val[v[i]]<val[ans]))ans=v[i],mx=t;

    }

    if(bl[a]!=bl[b])

        for(int i=(bl[b]-)*blo+; i<=b; i++) {

            int t=query(a,b,v[i]);

            if(t>mx||(t==mx&&val[v[i]]<val[ans]))ans=v[i],mx=t;

        }

    return ans;

}

int main() {

    n=read();

    blo=;

    for(int i=; i<=n; i++) {

        v[i]=read();

        if(!mp[v[i]]) {

            mp[v[i]]=++id;

            val[id]=v[i];

        }

        v[i]=mp[v[i]];

        ve[v[i]].push_back(i);

    }

    for(int i=; i<=n; i++)bl[i]=(i-)/blo+;

    for(int i=; i<=bl[n]; i++)pre(i);

    for(int i=; i<=n; i++) {

        int a=read(),b=read();

        if(a>b)swap(a,b);

        printf("%d\n",val[query(a,b)]);

    }

    return ;

}

分块算法小结：

　　暴力算法

　　时间复杂度可以

　　空间允许

　　优美

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