hdu5443 The Water Problem

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

 

Recommend

hujie   |   We have carefully selected several similar problems for you:  5449 5448 5447 5446 5445

区间求最大值

次裸裸的RMQ，套上模板即过

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<algorithm>
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 ;
}