The Water Problem（RMQ）

The Water Problem

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Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 1634 Accepted Submission(s): 1245

[align=left]点击打开链接[/align]
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.

[align=left]Input[/align]
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
[align=left]For each query, output an integer representing the size of the biggest water source.[/align]

[align=left]Sample Input[/align]

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;
}