The Water Problem(排序)

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：

#include<stdio.h>
#include<stdlib.h> 
#include<string.h>
const int MAXN=1010;
int cmp(const void *a,const void *b){
    if(*(int *)a<*(int *)b)return 1;
    else return -1;
}
int main(){
    int m[MAXN],T,N,q,l,r,n[MAXN];
    scanf("%d",&T);
    while(T--){
        scanf("%d",&N);
        for(int i=1;i<=N;i++)
            scanf("%d",m+i),n[i]=m[i];
        scanf("%d",&q);
        while(q--){
            scanf("%d%d",&l,&r);
            for(int i=1;i<=N;i++)
                m[i]=n[i];
            qsort(m+l,r-l+1,sizeof(m[0]),cmp);
            printf("%d\n",m[l]);
        }
    }
    return 0;
}