ST表模板 Balanced Lineup POJ3264

http://poj.org/problem?id=3264

题意 rmq max min之差

模板：

#define _CRT_SECURE_NO_WARNINGS
#include<cmath>
#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
#define rep(i,t,n)  for(int i =(t);i<=(n);++i)
#define per(i,n,t)  for(int i =(n);i>=(t);--i)
#define mmm(a,b) memset(a,b,sizeof(a))
const int maxn = 5e5 + 10;
int cow[maxn], minc[maxn][20], maxc[maxn][20];
int n, q;

void STinit() {
    int LOG;
    rep(i, 1, n)minc[i][0] = maxc[i][0] = cow[i];
    LOG = int(log(1.0*n) / log(1.0 * 2));
    rep(i,1,LOG)
        per(j, n, 1) {
        maxc[j][i] = maxc[j][i - 1];
        if (j + (1 << (i - 1)) <= n)
            maxc[j][i] = max(maxc[j][i], maxc[j + (1 << (i - 1))][i - 1]);
        minc[j][i] = minc[j][i - 1];
        if (j + (1 << (i - 1)) <= n)
            minc[j][i] = min(minc[j][i], minc[j + (1 << (i - 1))][i - 1]);
    }
}
int RMQmax(int l,int r) {
    int n = r - l + 1;
    int LOG = int(log(1.0*n) / log(1.0 * 2));
    return max(maxc[l][LOG], maxc[r -( 1 << LOG)+1][LOG]);
}
int RMQmin(int l, int r) {
    int n = r - l + 1;
    int LOG = int(log(1.0*n) / log(1.0 * 2));
    return min(minc[l][LOG], minc[r -( 1 << LOG)+1][LOG]);
}
int main() {
    cin >> n>>q;
    
    rep(i, 1, n)scanf("%d", &cow[i]);
    STinit();
    rep(i, 1, q) {
        int l, r;
        scanf("%d%d", &l, &r);
        
        printf("%d\n", RMQmax(l, r) - RMQmin(l, r));
    }
    cin >> n;
    return 0;
}