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

题意,给出一串数字,有m个询问,询问l-r之间最大值和最小值的差.

好吧,这明显可以用线段树做....我知道....但可以离线实在让我忍不住了....不管了....st做法奉上....

于是,收到了素质RE四连...

是是是,我以为网卡了,然后手贱直接交了4发,仔细一查原来j没算对,开大了....

然后改一发A掉了

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<cmath>

using namespace std;

int dp1[][],dp2[][],a[];

int main()

{

    int n,m;

    scanf("%d%d",&n,&m);

    for(int i=;i<=n;i++)

    {

        scanf("%d",&a[i]);

        dp1[i][]=a[i];

        dp2[i][]=a[i];

    }

    for(int j=;j<=;j++)

    {

        for(int i=;i<=n;i++)

        {

            if(i+(<<j)-<=n)

            dp1[i][j]=max(dp1[i][j-],dp1[i+(<<(j-))][j-]);

            dp2[i][j]=min(dp2[i][j-],dp2[i+(<<(j-))][j-]);

        }

    }

    for(int i=;i<=m;i++)

    {

        int l,r,ans=;

        scanf("%d%d",&l,&r);

        int x=(int)(log((double)(r-l+))/log(2.0));

        int max1=max(dp1[l][x],dp1[r-(<<x)+][x]);

        int min1=min(dp2[l][x],dp2[r-(<<x)+][x]);

        ans=max1-min1;

        printf("%d\n",ans);

    }

}