题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5443
The Water Problem
Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1084 Accepted Submission(s): 863
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,...,anrepresenting
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
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题目大意：给出n个水池的水量，找出区间内最大的水量。
求区间最值。正常都採用线段树的方法。可是这题数据量不大。全部暴力就过了~
详见代码。
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int t;
int a[1010],l,r;
scanf("%d",&t);
while (t--)
{
int n;
scanf("%d",&n);
for (int i=1; i<=n; i++)
{
scanf("%d",&a[i]);
}
int q;
scanf("%d",&q);
while (q--)
{
int Max=0;
scanf("%d%d",&l,&r);
for (int i=l; i<=r; i++)
{
if (a[i]>Max)
Max=a[i];
}
cout<<Max<<endl;
}
}
return 0;
}
