The Water Problem

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 816    Accepted Submission(s): 657

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

代码：

1 #include
       
         2 #include
        
           3 #include
         
           4 const int MAXN=1010; 5 int cmp(const void *a,const void *b){ 6     if(*(int *)a<*(int *)b)return 1; 7     else return -1; 8 } 9 int main(){10     int m[MAXN],T,N,q,l,r,n[MAXN];11     scanf("%d",&T);12     while(T--){13         scanf("%d",&N);14         for(int i=1;i<=N;i++)15             scanf("%d",m+i),n[i]=m[i];16         scanf("%d",&q);17         while(q--){18             scanf("%d%d",&l,&r);19             for(int i=1;i<=N;i++)20                 m[i]=n[i];21             qsort(m+l,r-l+1,sizeof(m[0]),cmp);22             printf("%d\n",m[l]);23         }24     }25     return 0;26 }