士兵杀敌 三 【ST算法】

题目链接：
http://acm.nyist.net/JudgeOnline/problem.php?pid=119

解法：　RMQ问题。

代码：

#include <stdio.h>
#include <ctime>
#include <math.h>
#include <limits.h>
#include <complex>
#include <string>
#include <functional>
#include <iterator>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <bitset>
#include <sstream>
#include <iomanip>
#include <fstream>
#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <time.h>
#include <ctype.h>
#include <string.h>
#include <assert.h>

using namespace std;

int n, k, p, l, r;

int a[100010];
int s[100010][20];
int maxsum[100010][20], minsum[100010][20];

int rmq(int l, int r)
{
    int k = log2((double)(r - l + 1));
    int MAX = max(maxsum[l][k], maxsum[r - (1 << k) + 1][k]);
    int MIN = min(minsum[l][k], minsum[r - (1 << k) + 1][k]);
    return MAX - MIN;
}

int main()
{
    int t;
    int q;
    while (scanf("%d %d", &n, &q)!=EOF)
    { 
        for (int i = 1; i <= n; i++)
        {
            scanf("%d", &a[i]); 
            maxsum[i][0] = minsum[i][0] = a[i];
        }

        for (int j = 1; (1 << j) <= n; j++)
            for (int i = 1; i + (1 << j) - 1 <= n; i++)
                {
                    maxsum[i][j] = max(maxsum[i][j - 1], maxsum[i + (1 << (j - 1))][j - 1]);
                    minsum[i][j] = min(minsum[i][j - 1], minsum[i + (1 << (j - 1))][j - 1]);
                }

        int s, t;
        while (q--)
        {
            scanf("%d%d",&s,&t);
            printf("%d\n", rmq(s, t));
        }
    }
    return 0;
}