LibreOJ-10119RMQ (Range Minimum / Maximum Query) problem

Enter a string of numbers and give you  MM queries. Each query will give you two numbers  X , Y X, Y, asking you to say  the maximum number in the interval X X to  Y Y.

Input

In the first line, two integers  N , M N, M represent the number of digits and the number of inquiries; the 
next line is  N N numbers; in the 
next  M M line, each line has two integers  X , Y X, Y.

Output

The output is  M M lines, and each line outputs a number.

Example

Sample input

10 2
3 2 4 5 6 8 1 2 9 7
1 4
3 8

Sample output

5
8

Hint

For all data, 1 ≤ N ≤ 10 5 , 1 ≤ M ≤ 10 6 , 1 ≤ X ≤ Y ≤ N 1≤N≤105, 1≤M≤106, 1≤X≤Y≤N. The number does not exceed  C/C++ the  int range.

The main methods and complexity are as follows:

1. Simple (that is, search), O (n) -O (qn) online.

2. Line segment tree , O (n) -O (qlogn) online.

3, ST (in essence, dynamic programming ), O (nlogn) -O ( q) online.

ST algorithm (Sparse Table), taking the maximum value as an example, let d [i, j] represent the maximum value in the interval [i, i + 2 ^ j-1], then when asked about the interval [a, b] The maximum value of the answer is max (d [a, k], d [b-2 ^ k + 1, k]), where k is the largest one that satisfies 2 ^ k <= b-a + 1 (that is, length) k, that is, k = [ln (b-a + 1) / ln (2)].

The method of seeking d may be dynamic programming , d [i, j] = max (d [i, j-1], d [i + 2 ^ (j-1), j-1]).

4. RMQ standard algorithm: first reduce to LCA (Lowest Common Ancestor), then reduce to constrained RMQ, O (n) -O (q) online.

First, according to the original number of columns , the establishment of Cartesian tree , so that the problem of linear time statute of the LCA problem. The LCA problem can be reduced to RMQ in linear time, that is, the RMQ problem where the difference between any two adjacent numbers in the sequence is +1 or -1. There are constraints RMQ O (n) -O (1) solution line, so that the whole algorithm time complexity is O (n) -O (1) .

#include <bits/stdc++.h>
using namespace std;
inline int read(){
    char ch=getchar();int res=0,f=1;
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') res=res*10+ch-'0', ch =getchar();
    return res*f;
}
inline void write(int zx){
    if(zx<0) zx=-zx,putchar('-');
    if(zx<10) putchar(zx+'0');
    else{
        write(zx/10);
        putchar(zx%10+'0');
    }
}
int n,m,a[100010],f[100010][20];
void ST(){
    for(int i=1;i<=n;i++) f[i][0]=a[i];
    for(int j=1;(1<<j)<=n;j++){
        for(int i=1;i+(1<<j)-1<=n;i++){
            f[i][j]=max(f[i][j-1],f[i+(1<<(j-1))][j-1]);
        }
    }
}
int RMQ(int l,int r){
    int k=0;
    while((1<<(k+1))<=r-l+1) k++;
    return max(f[l][k],f[r-(1<<k)+1][k]);
}
int main(){
    n=read();m=read();
    for(int i=1;i<=n;i++) a[i]=read();
    ST();
    for(int i=1;i<=m;i++){
        int l=read(),r=read();
        write(RMQ(l,r));
        putchar('\n');
    }
    return 0;
}