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

区间最大值   线段树

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           using namespace std;const int oo = 0x3f3f3f3f;const int N = 1005;typedef long long LL;struct da{    int left, right, val;}as[N*4];int n, ac[N];void build(int left, int right, int i){    as[i].left = left;    as[i].right = right;    int mid = (left + right)/2;    if(left == right)    {        as[i].val = ac[left];        return ;    }    build(left, mid, 2*i);    build(mid+1, right, 2*i+1);    as[i].val = max(as[i*2].val, as[2*i+1].val);}void query(int left, int right, int i, int &ans){    if(as[i].left == left && as[i].right == right)    {        ans = max(as[i].val, ans);        return ;    }    int mid = (as[i].left + as[i].right)/2;    if(right <= mid)  query(left, right, 2*i, ans);    else if(left > mid) query(left, right, 2*i+1, ans);    else    {        query(left, mid, 2*i, ans);        query(mid+1, right, 2*i+1, ans);    }}int main(){    int T, Q;    scanf("%d", &T);    while(T--)    {        scanf("%d", &n);        for(int i = 1; i <= n; i++)            scanf("%d", &ac[i]);        build(1, n, 1);        scanf("%d", &Q);        while(Q--)        {            int a, b;            scanf("%d %d", &a, &b);            int ans = 0;            query(a, b, 1, ans);            printf("%d\n", ans);        }    }    return 0;}