题号未知 ， Array Queries 【RMQ】

Description

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 index I 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

题目大意：区间找最小值， 这里多一个操作， 就是对k进行了定位；确定在哪层 ，然后得到最小值； 
AC代码：

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
#define N 100000+10
int dp
[20],a
;
void initrmq(int n)
{
int i,j;
for(i=1;i<=n;i++)
dp[i][0]=a[i];
for(j=1;(1<<j)<=n;j++)
for(i=1;i+(1<<j)-1<=n;i++)
dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}
int rmq(int l,int r)
{
int k=(int)(log(r-l+1)/log(2.0));//*一共20层，定位到那层直接进行查找最小值
return min(dp[l][k],dp[r-(1<<k)+1][k]);
}
int main()
{
int n,m,i,cas=1,t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
initrmq(n);
printf("Case %d:\n",cas++);
while(m--)
{
int x , y ;
scanf("%d%d",&x,&y);
printf("%d\n",rmq(x,y));
}
}
return 0 ;
}