RMQ区间最大值【模板题】

模板1

51nod题目链接

• 区间dp
• 2的幂次

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>


using namespace std;
const int N = 2e5+5, M = 18;
int f[N][M], w[N];
int n;
void init(){
    
    
    //区间dp
    for(int j = 0;  j < M; j++){
    
    
        for(int i = 1; i + (1<<j) - 1 <= n; i++){
    
    
            if(!j) f[i][j] = w[i];
            else{
    
    
                f[i][j] = max(f[i][j-1], f[i + (1 << (j-1))][j-1]);
            }
        }
    }
}

int query(int l, int r){
    
    
    int len = r - l + 1;
    int k = log(len)/log(2);
    return max(f[l][k], f[r - (1<<k)+1][k]);
}
int main()
{
    
    
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d", &w[i]);
    init();
    int m;
    scanf("%d", &m);
    while (m -- ){
    
    
        int l, r; 
        scanf("%d%d", &l, &r);
        printf("%d\n", query(l, r));
    }
    return 0;
}

模板2

AcWing题目链接

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>


using namespace std;
const int N = 2e5+5, M = 18;
int f[N][M], w[N];
int n;
void init(){
    
    
    //区间dp
    for(int j = 0;  j < M; j++){
    
    
        for(int i = 1; i + (1<<j) - 1 <= n; i++){
    
    
            if(!j) f[i][j] = w[i];
            else{
    
    
                f[i][j] = max(f[i][j-1], f[i + (1 << (j-1))][j-1]);
            }
        }
    }
}

int query(int l, int r){
    
    
    int len = r - l + 1;
    int k = log(len)/log(2);
    return max(f[l][k], f[r - (1<<k)+1][k]);
}
int main()
{
    
    
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d", &w[i]);
    init();
    int m;
    scanf("%d", &m);
    while (m -- ){
    
    
        int l, r; 
        scanf("%d%d", &l, &r);
        printf("%d\n", query(l, r));
    }
    return 0;
}