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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 Output for Sample Input Case 1: 1 3 12 Case 2: 10

很裸的线段树

#include 
   
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            #define INF 0x3f3f3f3f#define MAXN 25#define Mod 10001using namespace std;const int maxn=222222;int min(int a,int b){ if(a
            
             >1; build(l,m,rt<<1); build(m+1,r,rt<<1|1); PushUp(rt);}int query(int L,int R,int l,int r,int rt){ if(L<=l&&r<=R) return MIN[rt]; int m=(l+r)>>1; int ret=INF; if(L<=m) ret=min(ret,query(L,R,l,m,rt<<1)); if(R>m) ret=min(ret,query(L,R,m+1,r,rt<<1|1)); return ret;}int main(){ int t; scanf("%d",&t); for(int cas=1; cas<=t; ++cas) { int n,q,l,r; scanf("%d%d",&n,&q); build(1,n,1); printf("Case %d:\n",cas); while(q--) { scanf("%d%d",&l,&r); printf("%d\n",query(l,r,1,n,1)); } } return 0;}
            
           
          
         
        
       
      
     
    
   

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