无题

 

问题

Given an array with N elements, indexed from 1 to N. Now you will be given some queries in the form I J, your task is to find the minimum value from index I to J.

Input

Input starts with an integer T (≤ 5), denoting the number of test cases.

The first line of a case is a blank line. The next line contains two integers N (1 ≤ N ≤ 10⁵), q (1 ≤ q ≤ 50000). The next line contains N space separated integers forming the array. There integers range in [0, 10⁵].

The next q lines will contain a query which is in the form I J (1 ≤ I ≤ J ≤ N).

Output

For each test case, print the case number in a single line. Then for each query you have to print a line containing the minimum value between indexI and J.

Sample Input

2

 

5 3

78 1 22 12 3

1 2

3 5

4 4

 

1 1

10

1 1

Sample Output

Case 1:

1

3

12

Case 2:

10

//这是超时的

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std;int a[100100],b[100100];int main(){int t,T=0;scanf("%d",&t);while(t--){int n,m,x,y;scanf("%d%d",&n,&m);for(int i=1;i<=n;i++)scanf("%d",&a[i]);printf("Case %d:\n",++T);while(m--){int j=1;for(int i=1;i<=n;i++)b[j++]=a[i];scanf("%d%d",&x,&y);sort(b+x,b+y+1);printf("%d\n",b[x]);}}return 0;}