Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 53721		Accepted: 25244
Case Time Limit: 2000MS

For the daily milking, Farmer John‘s N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.

Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.

Line 1: Two space-separated integers, N and Q.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i
Lines N+2..N+Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.

Lines 1..Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.

6 3
1
7
3
4
2
5
1 5
4 6
2 2

6
3
0

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<string.h>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<cmath>
using namespace std;
#define N 300000
struct node{
    int left,right,maxx,minn;
}tree[N*4];
int n,m,ans;
void build(int l,int r,int pos){                         //建树
    tree[pos].left=l;
    tree[pos].right=r;
    tree[pos].maxx=0;                                    //。。。
    tree[pos].minn=1e9;                                  //一开始写的20000是不对的！！！看清题意。。。
    if(tree[pos].left==tree[pos].right)return;
    int mid=(l+r)>>1;
    build(l,mid,pos*2);
    build(mid+1,r,pos*2+1);
}
void update(int x,int y,int pos){
    if(tree[pos].left==x&&tree[pos].right==x){
        tree[pos].maxx=y;
        tree[pos].minn=y;
        return;
    }
    int mid=(tree[pos].left+tree[pos].right)>>1;
    if(x>mid)
        update(x,y,pos*2+1);
    else
        update(x,y,pos*2);
  tree[pos].maxx=max(tree[pos*2].maxx,tree[pos*2+1].maxx);
  tree[pos].minn=min(tree[pos*2].minn,tree[pos*2+1].minn);
}
int query1(int x,int y,int pos){                          //查询最大值          
    if(tree[pos].left==x&&tree[pos].right==y){
        ans=max(ans,tree[pos].maxx);
        return ans;
    }
    int mid=(tree[pos].left+tree[pos].right)>>1;
    if(y<=mid)query1(x,y,pos*2);
    else if(x>mid)query1(x,y,pos*2+1);
    else{
        query1(x,mid,pos*2);
        query1(mid+1,y,pos*2+1);
    }
}
int query2(int x,int y,int pos){                          //查询最小值
    if(tree[pos].left==x&&tree[pos].right==y){
        ans=min(ans,tree[pos].minn);
        return ans;
    }
    int mid=(tree[pos].left+tree[pos].right)>>1;
    if(y<=mid)query2(x,y,pos*2);
    else if(x>mid)query2(x,y,pos*2+1);
    else{
        query2(x,mid,pos*2);
        query2(mid+1,y,pos*2+1);
    }
}
int main(){
    int m,n;
    int ans1,ans2;
    while(~scanf("%d%d",&n,&m)){
        getchar();
        build(1,n,1);
        int x;
        for(int i=1;i<=n;i++){
            scanf("%d",&x);
            update(i,x,1);
        }
        int a,b;
        while(m--){
            scanf("%d%d",&a,&b);
                ans=0;
                ans1=query1(a,b,1);
                ans=1e9;                              //一开始写的200000，是不对的！！！
                ans2=query2(a,b,1);
                printf("%d\n",ans1-ans2);
        }
    }
    return 0;
}

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<string.h>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<cmath>
using namespace std;
const int N=2*1e5+10;
int a[N],b[N],c[N];

int main(){
    int n,m,h,k;
    int minn,maxx,ans;
    while(~scanf("%d%d",&n,&m)){
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
            for(int i=0;i<n;i++)
                b[i]=a[i];                           //b是数按顺序排
            sort(b,b+n);
            for(int i=0;i<n;i++){
                c[b[i]]=i;                           //c是数对应的第几大
            }
        for(int i=0;i<m;i++){
            scanf("%d%d",&h,&k);
            minn=N;maxx=-1;
            for(int i=h-1;i<=k-1;i++){

                minn=min(minn,c[a[i]]);
                maxx=max(maxx,c[a[i]]);
            }
            ans=b[maxx]-b[minn];
            printf("%d\n",ans);
        }
    }
    return 0;
}

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<string.h>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<cmath>
using namespace std;
const int N=2*1e5+10;
int a[N],b[N],c[N],d[N];

int main(){
    int n,m,h,k;
    int minn,maxx,ans;
    while(~scanf("%d%d",&n,&m)){
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
            for(int i=0;i<n;i++)
                b[i]=a[i];                           //b是数按顺序排
            sort(b,b+n);
            for(int i=0;i<n;i++){
                c[b[i]]=i;                           //c是数对应的第几大
            }
        for(int i=0;i<m;i++){
            scanf("%d%d",&h,&k);
            minn=N;maxx=-1;
            int j=0;
            for(int i=h-1;i<=k-1;i++){
                    d[j++]=c[a[i]];
                    sort(d,d+j);
                    minn=d[0];
                    maxx=d[j-1];
                //minn=min(minn,c[a[i]]);
                //maxx=max(maxx,c[a[i]]);
            }
            ans=b[maxx]-b[minn];
            printf("%d\n",ans);
        }
    }
    return 0;
}