[NYOJ 119] 士兵杀敌（三） （RMQ ST算法）

/*
ID: wuqi9395@126.com
PROG: beads
LANG: C++
*/
#include<iostream>
#include<fstream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<ctype.h>
#include<algorithm>
#include<string>
#define PI acos(-1.0)
#define maxn 100010
#define INF 1<<25
#define mem(a, b) memset(a, b, sizeof(a))
typedef long long ll;
using namespace std;
int N, Q, mx[20][maxn], mn[20][maxn];
void RMQ(int num)
{
    for (int i = 1; i <= log2(num) + 1; i++)
        for (int j = 1; j <= num; j++) if (j + (1 << i) - 1 <= num)
        {
            mx[i][j] = max(mx[i - 1][j], mx[i - 1][j + (1 << i >> 1)]);
            mn[i][j] = min(mn[i - 1][j], mn[i - 1][j + (1 << i >> 1)]);
        }
}
int main ()
{
    scanf("%d%d", &N, &Q);
    for (int i = 1; i <= N; i++)
    {
        scanf("%d", &mx[0][i]);
        mn[0][i] = mx[0][i];
    }
    RMQ(N);
    int st, ed;
    while(Q--)
    {
        scanf("%d%d", &st, &ed);
        int k = (int) log2(ed - st + 1.0);
        int mxans = max(mx[k][st], mx[k][ed - (1 << k) + 1]);
        int mnans = min(mn[k][st], mn[k][ed - (1 << k) + 1]);
        printf("%d\n", mxans - mnans);
    }
    return 0;
}