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
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1005;
int a[maxn];int n;
int h[maxn];
int lowbit(int x){
return x&(-x);
}
void update(int i,int val){
while(i<=n){
h[i]=max(h[i],val);
i+=-i&i;
}
}
int find(int l,int r)
{
int ret=a[r];
while(1)
{
ret=max(ret,a[r]);
if(l==r)break;
for(r-=1;r-l>=lowbit(r);r-=lowbit(r))
ret=max(ret,h[r]);
}
return ret;
}
int query(int x,int y){//这个查询不正确
int ans=a[y];
while(y>=x){
ans=max(a[y],ans);
y--;
for(;y-lowbit(y)>=x;y-=lowbit(y)){
ans=max(h[x],ans);
}
}
return ans;
}
int main(){
int T;cin>>T;
while(T--){
cin>>n;
memset(a,0,sizeof(a));
memset(h,0,sizeof(h));
for(int i=1;i<=n;i++){
cin>>a[i];
update(i,a[i]);
}
int q;cin>>q;
int l,r;
while(q--){
cin>>l>>r;
cout<<find(l,r)<<endl;
}
}
return 0;
}