• dp[i][j]表示區間[i,i+2j)的最小值
• dp[i][j]=min(dp[i][j−1],dp[i+2j−1][j−1]
• 對於詢問[le,ri]，令len=ri−le+1，再令2lb≤len<2lb+1，則min[le,ri]=min{dp[le][lb],dp[ri−2lb+1][lb]}

/* **********************************************

    File Name: 1068.cpp

    Auther: [email protected]

    Created Time: 2015/9/25 星期五 下午 9:19:08

*********************************************** */
 

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const int MAX = 1000007;
const int MAX_L = 21;
int dp[MAX][MAX_L];
int a[MAX];
int po[MAX_L];

int main() {
    po[0] = 1;
    for (int i = 1; i < MAX_L; ++i) {
        po[i] = po[i - 1] << 1
 
;
    }

    int n;
    scanf(" %d", &n);
    for (int i = 1; i <= n; ++i) {
        scanf(" %d", a + i);
    }
    for (int i = n; i >= 1; --i) {
        dp[i][0] = a[i];
        for (int j = 1; i + (1 << j) <= n + 1; ++j) {
            //[i, i + (1 << j))
            //[i, i + (1 << (j - 1))) | [i + (1 << (j - 1)), i + (1 << j))
 

            //=>
            //dp[i][j - 1] | dp[i + (1 << (j - 1))][j - 1]
            dp[i][j] = min(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
            //printf("dp[%d][%d] = %d\n", i, j, dp[i][j]);
        }
    }

    int m;
    scanf(" %d", &m);
    while (m--) {
        int le, ri;
        scanf(" %d %d", &le, &ri);
        int lb = upper_bound(po, po + MAX_L, ri - le + 1) - po - 1;
        //printf("lb = %d\n", lb);
        //printf("cut to dp[%d][%d] and dp[%d][%d]\n", le, lb, ri - po[lb] + 1, lb);
        printf("%d\n", min(dp[le][lb], dp[ri - po[lb] + 1][lb]));
    }
    return 0;
}