The Water Problem（RMQ）

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The Water Problem

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1634 Accepted Submission(s): 1245

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Problem Description

In Land waterless, water is a very limited resource. People always fight for the biggest source of water. Given a sequence of water sources with

a1,a2,a3,...,an representing the size of the water source. Given a set of queries each containing

2 integers

l and

r , please find out the biggest water source between

al and

ar .

Input

First you are given an integer

T(T≤10) indicating the number of test cases. For each test case, there is a number

n(0≤n≤1000) on a line representing the number of water sources.

n integers follow, respectively

a1,a2,a3,...,an , and each integer is in

{1,...,106} . On the next line, there is a number

q(0≤q≤1000) representing the number of queries. After that, there will be

q lines with two integers

l and

r(1≤l≤r≤n) indicating the range of which you should find out the biggest water source.

Output

For each query, output an integer representing the size of the biggest water source.

Sample Input

3
1
100
1
1 1
5
1 2 3 4 5
5
1 2
1 3
2 4
3 4
3 5
3
1 999999 1
4
1 1
1 2
2 3
3 3

Sample Output

100
2
3
4
4
5
1
999999
999999
1

题源：2015亚洲赛区长春网络赛

题目分析：区间求最大值。RMQ，模板题。

AC代码：

#include <iostream>
#include <cmath>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#define maxn 10005
using namespace std;
int maxSum[maxn][20];

void RMQ(int num)    //预处理->O(nlogn)
{
   for(int j = 1;j < 20;j++)
   {
       for(int i = 1;i <= num;i++)
       {
           if((i + (1 << j) - 1) <= num)//区间长度要满足总数条件
           {
               maxSum[i][j] = max(maxSum[i][j - 1],maxSum[i + (1 << (j - 1))][j - 1]);
           }
       }
   }
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
        {
            int num,query;
        int src,des;
        scanf("%d",&num);
        for(int i = 1;i <= num;i++)
        {
            scanf("%d",&maxSum[i][0]);//表示从第i个数起连续2^0（1）个数，即就是这第i个数
        }
        RMQ(num);
        scanf("%d",&query);
        while(query--)//O(1)查询
        {
            scanf("%d %d",&src,&des);
            int k = (int)(log(des - src + 1.0) / log(2.0));//中间位置
            int ansMax = max(maxSum[src][k],maxSum[des - (1 << k) + 1][k]);
            printf("%d\n",ansMax);
        }
    }

    return 0;
}