POJ 3264

RMQ模板题

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <string.h>
 4 #include <cmath>
 5 using namespace std;
 6 const int N= 50002;
 7 int maxn[N][17],minn[N][17];
 8 int a[N];
 9 int n,q;
10 void getRMQ(){
11     for(int i=1;i<N;i++){maxn[i][0]=minn[i][0]=a[i];}
12     for(int j=1;j<17;j++){
13         for(int i=1;i<N;i++){
14             if(i+(1<<j)<N){
15                 maxn[i][j]=max(maxn[i][j-1],maxn[i+(1<<(j-1))][j-1]);
16                 minn[i][j]=min(minn[i][j-1],minn[i+(1<<(j-1))][j-1]);
17             }
18         }
19     }
20 }
21 int rmq(int l,int r){
22     int k=(int)(log((double)(r-l+1))/log(2.0));
23     int Max=max(maxn[l][k],maxn[r-(1<<k)+1][k]);
24     int Min=min(minn[l][k],minn[r-(1<<k)+1][k]);
25     return Max-Min;
26 }
27 int main(){
28     int l,r;
29     scanf("%d%d",&n,&q);
30     for(int i=1;i<=n;i++){
31         scanf("%d",&a[i]);
32     }
33     getRMQ();
34     while(q--){
35         scanf("%d%d",&l,&r);
36         printf("%d\n",rmq(l,r));
37     }
38     return 0;
39 }