[poj3264]rmq算法学习(ST表)

解题关键：rmq模板题，可以用st表，亦可用线段树等数据结构

log10和log2都可，这里用到了对数的换底公式

类似于区间dp，用到了倍增的思想

$F[i][j] = \min (F[i][j - 1],F[i + 1 <  < (j - 1)][j - 1])$

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<iostream>
using namespace std;
typedef long long ll;
int a[50002];
int lg;
int min1[50002][22],max1[50002][22],n,q;
void rmq(int n){
    for(int j=1;j<=lg;j++){
        for(int i=1;i+(1<<j)-1<=n;i++){
            max1[i][j]=max(max1[i][j-1],max1[i+(1<<(j-1))][j-1]);
            min1[i][j]=min(min1[i][j-1],min1[i+(1<<(j-1))][j-1]);
        }
    }
}
int main(){
    int n,q;
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin>>n>>q;
    lg=int(log10(n)/log10(2));
    for(int i=1;i<=n;i++) cin>>a[i],min1[i][0]=max1[i][0]=a[i];
    rmq(n);
    while(q--){
        int a,b;
        cin>>a>>b;
        if(a>b) swap(a,b);
        int k=(int)(log10(b-a+1)/log10(2.0));
        int maxres=max(max1[a][k],max1[b-(1<<k)+1][k]);
        int minres=min(min1[a][k],min1[b-(1<<k)+1][k]);
        cout<<maxres-minres<<"\n";
    }
}