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st表

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st表

询问特定区间的最大/最小值

先进行初始化，每次询问的复杂度为O(1)

询问区间的极差(最大值-最小值)

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/*
 * @Description: 
 * @Author: CrayonXiaoxin
 * @Date: 2024-01-26 10:46:31
 * @LastEditTime: 2024-05-22 20:53:39
 * @LastEditors:  
 */
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
const int maxn = 1e6 + 10;
int n, m;
ll max_xy[maxn][22], min_xy[maxn][22];

ll st(int l,int r)
{
    int s = log2(r - l + 1), x, y;
    x = max(max_xy[l][s], max_xy[r - (1 << s) + 1][s]);//区间最大
    y = min(min_xy[l][s], min_xy[r - (1 << s) + 1][s]);//区间最小
    return x - y;
}
void solve(){
    cin >> n >> m;
    for (int i = 1; i <= n;++i)
    {
        cin >> max_xy[i][0];
        min_xy[i][0] = max_xy[i][0];
    }

    for (int i = 1; i <= 21;++i)//这个循环的上界决定于数据的大小，即2的21次方大于数据
    {
        for (int k = 1; k + (1 << i) <= n + 1;++k)
        {
            max_xy[k][i] = max(max_xy[k][i - 1], max_xy[k + (1 << (i - 1))][i - 1]);
            min_xy[k][i] = min(min_xy[k][i - 1], min_xy[k + (1 << (i - 1))][i - 1]);
        }
    }

    while(m--)
    {
        int l, r;
        cin >> l >> r;
        cout << st(l, r) << endl;
    }
}

int main(){
//std::ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
int t=1;
//cin>>t;
while(t--){
    solve();
}
return 0;
}