连模板都不会了丢脸。
思路:设f[i][j]为从i~i+2j的最大值,然后就是正常的RMQ了qwq,挂在36pts.
code:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <stdlib.h>
#include <stack>
#include <queue>
#define ri register int
#define N 1000005
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
return x*f;
}
int f[N][50],n,m;
int a[N];
int query(int l,int r){
if(l==r)return a[l];
int k=log2(r-l);
return std::max(f[l][k],f[r-(1<<k)][k]);
}
int main(){
n=read(),m=read();
for(ri i=1;i<=n;i++)a[i]=read();
for(ri i=1;i<n;i++)f[i][0]=std::max(a[i],a[i+1]);
for(ri k=1;k<=15;k++){
for(ri i=1;i+(1<<k)<=n;i++){
f[i][k]=std::max(f[i][k-1],f[i+(1<<(k-1))][k-1]);
}
}
for(ri i=1;i<=m;i++){
int l=read(),r=read();
printf("%d\n",query(l,r));
}
return 0;
}