The Water Problem

**Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 602    Accepted Submission(s): 485
**

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

Source

2015 ACM/ICPC Asia Regional Changchun Online

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区间求最大值

次裸裸的RMQ，套上模板即过

#include
#include
#include
#include
#include
using namespace std;
#define N 50005
int dpmin[N][],dpmax[N][];
int main()
{
int t;
scanf("%d",&t);
while(t--){
int i, j, n, m;
scanf("%d",&n);
memset(dpmin,,sizeof(dpmin));
memset(dpmax,,sizeof(dpmax));
for(i=; i<=n; i++)
{
scanf("%d", &dpmin[i][]);
dpmax[i][]=dpmin[i][];
}
int mm=floor(log(1.0*n)/log(2.0));
for(j=; j<=mm; j++)
for(i=; i<=n; i++)
{
if((i+(<<(j-)))<=n)
{
// dpmin[i][j]=min(dpmin[i][j-1], dpmin[i+(1<<(j-1))][j-1]);
dpmax[i][j]=max(dpmax[i][j-], dpmax[i+(<<(j-))][j-]);
}
}
int x, y;

    scanf("%d",&m);  
for(int i=; i<=m; i++)  
{  
    scanf("%d%d", &x, &y);  
    int  mid=floor(log(y\*1.0-x+)/log(2.0));  
    int maxnum=max(dpmax\[x\]\[mid\], dpmax\[y-(<<mid)+\]\[mid\]);  
//    int minnum=min(dpmin\[x\]\[mid\], dpmin\[y-(1<<mid)+1\]\[mid\]);  
    printf("%d\\n", maxnum);  
}  
}  
return ;  

}