POJ 3264 Balanced Lineup

/*
RMQ问题——稀疏表算法
状态转移方程
dp[i,j]=min{dp[i,j-1],dp[i+2j-1,j-1]}
*/
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
const int MAXN = 50001;
int max[MAXN][16],min[MAXN][16];    //2^16 = 65536
int a[MAXN];
int n, Q;
inline int getMax(int a, int b){
    return a > b ? a : b;
}
inline int getMin(int a, int b){
    return a < b ? a : b;
}
void RMQ_init(){
    int i, j, m;
    for(i = 1; i <= n; i++)
        max[i][0] = min[i][0] = a[i];    //i ~ i*2^0 即a[i]
    m = (int)(log((double)n)/log(2.0));
    for(j = 1; j <= m; ++j){        //自底向上递推
        for(i = n; i > 0; i--){
            max[i][j] = getMax(max[i][j - 1], max[i + (1 << (j - 1))][j - 1]);
            min[i][j] = getMin(min[i][j - 1], min[i + (1 << (j - 1))][j - 1]);
        }
    }
}
int findMax(const int f,const int t){
    if( f == t )
        return max[f][0];
    int m = (int)(log((double)(t - f + 1))/log(2.0));
    return getMax(max[f][m], max[t + 1 - (1 << (m))][m]);
}
int findMin(const int f,const int t){
    if( f == t )
        return min[f][0];
    int m = (int)(log((double)(t - f + 1))/log(2.0));
    return getMin(min[f][m], min[t + 1 - (1 << (m))][m]);
}
int main(){
    int f, t;
    scanf("%d%d",&n, &Q);
    for(int i = 1; i <= n; ++i)
        scanf("%d",&a[i]);
    RMQ_init();
    for(int i = 0; i < Q; ++i){
        scanf("%d%d", &f, &t);
        if(f > t)
            f ^= t ^= f ^= t;
        printf("%d\n", findMax(f, t) - findMin(f, t)); 
    }
    return 0;
}