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,...,10^6} . 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
题目大意就是给定区间，求区间最值。
这里采用了RMQ的ST算法，直接套的模板。
代码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring> using namespace std; const int maxN = ;
int n, q;
int a[maxN];
int ma[maxN][]; void RMQ()
{
memset(ma, , sizeof(ma));
for (int i = ; i < n; ++i)
ma[i][] = a[i];
for (int j = ; (<<j) <= n; ++j)
for (int i = ; i+(<<j)- < n; ++i)
ma[i][j] = max(ma[i][j-], ma[i+(<<(j-))][j-]);
} int query(int lt, int rt)
{
int k = ;
while ((<<(k+)) <= rt-lt+)
k++;
return max(ma[lt][k], ma[rt-(<<k)+][k]);
} void input()
{
scanf("%d", &n);
for (int i = ; i < n; ++i)
scanf("%d", &a[i]);
RMQ();
} void work()
{
scanf("%d", &q);
int u, v, ans;
for (int i = ; i < q; ++i)
{
scanf("%d%d", &u, &v);
ans = query(u-, v-);
printf("%d\n", ans);
}
} int main()
{
//freopen("test.in", "r", stdin);
int T;
scanf("%d", &T);
for (int times = ; times < T; ++times)
{
input();
work();
}
return ;
}