洛谷P3865 ST表

传送门啦

思路：

$ f[i][j] $ 表示从 $ i $ 开始，包含 $ 1<<j $ 个元素的区间的区间最大值；

转移方程: $ f[i][j]=max_(f[i][j-1],f[i+(1<<j-1)][j-1] $ ;

查询 $ (l,r) $ :

$ p=log_2(r-l+1) $ ;

$ max(l,r)=max(f[l][p],f[r-(1<<p)+1][p]) $ ;

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath> 
#define re register
using namespace std ;

int n , m , a[100005] , l , r ;
int f[10000010][21] ;

inline int read () {
	int f = 1 , x = 0 ; 
	char ch = getchar () ;
	while(ch > '9' || ch < '0') {if(ch == '-') f = -1 ; ch = getchar () ;}
	while(ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + ch - '0' ; ch = getchar () ;}
	return x * f ;
}

inline void st(int x) {
	for(re int i = 1 ; i <= 21 ; ++ i) 
		for(re int j = 1 ; j + (1 << i) <= x + 1 ; ++ j)
			f[j][i] = max(f[j][i - 1] , f[j + (1 << (i -1))][i - 1]) ;
}

inline int query(int l , int r) {
	int k = log2(r - l + 1) ;
	return max(f[l][k] , f[r - (1 << k) + 1][k]) ;
}

int main () {
	n = read () ; m = read () ;
	for(re int i = 1 ; i <= n ; ++ i) {
		f[i][0] = read () ;
	}
	st(n) ;
	for(re int i = 1 ; i <= m ; ++ i) {
		l = read () ; r = read () ;
		printf("%d\n" , query(l , r)) ;
	}
	return 0 ;
}