線段樹好像過不了，而且還卡cin，cout -_-
ST演算法就是用二維陣列儲存當前位置的$2^j$的區間的最小值
而且由遞推公式 $dp[i][j] = min(dp[i][j - 1], dp[i + 2^(j - 1)][j - 1])$

那麼求出dp陣列的話，那麼就可解

AC程式碼：

c

#include <bits/stdc++.h>

using namespace std;

const int maxn =
 
 1e6 + 5;
const int Mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
#define LL long long

int dp[maxn][30];

int main()
{
    int n;
    scanf("%d", &n);
    memset(dp, INF, sizeof(dp));
    for (int i = 1; i <= n; i ++)
        scanf("%d", &dp[i][0]);
    int base = 1;
    for (int i = 1; (base <<
 
 1) <= n; i ++) {
        for (int j = 1; j + base <= n; j ++) {
            dp[j][i] = min(dp[j][i - 1], dp[j + base][i - 1]);
        }
        base <<= 1;
    }

    int q;
    scanf("%d", &q);
    while(q --) {
        int L, R;
        scanf("%d %d", &L, &R);
        int
 
 p = log2(R - L + 1), k = 1 << p;
        int ans = min(dp[L][p], dp[R - k + 1][p]);
        printf("%d\n", ans);
    }
    return 0;
}