【杭電】5443 水題

L - L Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u Submit Status

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 $a_1, a_2, a_3, . . . , a_n$ 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 $a_l$ and $a_r$.  

Input

First you are given an integer $T (T \leq 10)$ indicating the number of test cases. For each test case, there is a number $n (0 \leq n \leq 1000)$ on a line representing the number of water sources. $n$ integers follow, respectively $a_1, a_2, a_3, . . . , a_n$, and each integer is in $\{1, . . . , 10^6\}$. On the next line, there is a number $q (0 \leq q \leq 1000)$ representing the number of queries. After that, there will be $q$ lines with two integers $l$ and $r (1 \leq l \leq r \leq n)$ 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 
     
     
      code：
     
     
             
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
	int l,r,t;int a[1010],b[1010];
	scanf("%d",&t);
	while(t--){
		memset(b,0,sizeof(b));
		int n;
		scanf("%d",&n);
		for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
		int p;
		scanf("%d",&p);
		for(int i=1;i<=p;i++){
			scanf("%d%d",&l,&r);
			for(int j=l;j<=r;j++)
			b[j]=a[j];
			sort(b+l,b+r+1);
			printf("%d\n",b[r]);
		} 
	}
	return 0;
 }