poj 3264 RMQ

直接写个RMQ就能过。

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<cmath>

#define Max(a,b) ((a)>(b)?(a):(b))

#define Min(a,b) ((a)<(b)?(a):(b))

#define Maxn 60010

using namespace std;

int maxnum[Maxn][20],minnum[Maxn][20],n,Log[Maxn+10];

int Log2(int x)

{

    int num=0;

    x/=2;

    while(x)

    {

        num++;

        x/=2;

    }

    return num;

}

void RMQ()

{

    int i,j;

    for(j=1;j<20;j++)

        for(i=1;i<=n;i++)

        if(i+(1<<j)-1<=n)

        {

            maxnum[i][j]=Max(maxnum[i][j-1],maxnum[i+(1<<(j-1))][j-1]);

            minnum[i][j]=Min(minnum[i][j-1],minnum[i+(1<<(j-1))][j-1]);

        }

}

int main()

{

    int i,j,q,a,b;

    for(i=1;i<=60010;i++)

        Log[i]=Log2(i);

    while(scanf("%d%d",&n,&q)!=EOF)

    {



        for(i=1;i<=n;i++)

        {

            scanf("%d",&a);

            maxnum[i][0]=minnum[i][0]=a;

        }

        RMQ();

        for(i=1;i<=q;i++)

        {

            scanf("%d%d",&a,&b);

            int k=Log[b-a+1];

            printf("%d\n",Max(maxnum[a][k],maxnum[b-(1<<k)+1][k])-Min(minnum[a][k],minnum[b-(1<<k)+1][k]));

        }

    }

    return 0;

}