RMQ问题 - ST表的简单应用

/*
@theme:poj 3264
@writer:pprp
@declare:ST表（sparse table）稀疏表,用动态规划的思想来解决RMQ问题；
@date:2017/8/26
*/

//#include <bits/stdc++.h>
#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#define IOS ios::sync_with_stdio(false),cin.tie(0);

using namespace std;
const int maxn = 50050;

int F1[maxn][20];
int F2[maxn][20];
int a[maxn];

void SparseTable(int a[], int len)
{
    //初始化
    for(int i = 0 ; i < len ; i++)
    {
        F1[i][0] = a[i];
        F2[i][0] = a[i];
    }

    //递推
    //找到j的范围log2(n)
    int nlog = int(log(double(len))/log(2.0));
    for(int j = 1 ; j <= nlog; j++)
    {
        for(int i = 0 ; i < len ; i++)
        {
            //区间右端点不能超过数组最后一位下标
            if((i + (1 << j) -1) < len )
            {
                F1[i][j] = min(F1[i][j - 1],F1[i + (1 << (j - 1))][j - 1]);
                F2[i][j] = max(F2[i][j - 1],F2[i + (1 << (j - 1))][j - 1]);
            }
        }
    }
}

int main()
{
    int N, Q;
    int l, r;

    scanf("%d %d",&N,&Q);

    for(int i = 0 ; i < N ; i++)
    {
        scanf("%d",&a[i]);
    }

    SparseTable(a,N);

    for(int i = 0 ; i < Q ; i++)
    {
        scanf("%d %d",&l,&r);
        l--;
        r--;
        double len = r - l + 1;
        int m = (int)(log(len)/log(2.0));
        int mmx = max(F2[l][m],F2[r-(1<<m)+1][m]);
        int mmn = min(F1[l][m],F1[r-(1<<m)+1][m]);
        printf("%d\n",mmx-mmn);
    }


    return 0;
}