[NYOJ 119] attackers' competitor (3) (rmq st algorithm)

There are N soldiers in a row from left to right, numbered from 1 to N. Each of them has killed different numbers of enemies. Now we have Q queries. Each time we ask, we can find the difference between the range [L, R], which has the least number of attacks and the number of attacks. Rmq st algorithm bare question.

/*ID: wuqi9395@126.comPROG: beadsLANG: C++*/#include
 
  #include
  
   #include
   
    #include
    
     #include
     
      #include
      #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include#include
            
             #define PI acos(-1.0)#define maxn 100010#define INF 1<<25#define mem(a, b) memset(a, b, sizeof(a))typedef long long ll;using namespace std;int N, Q, mx[20][maxn], mn[20][maxn];void RMQ(int num){ for (int i = 1; i <= log2(num) + 1; i++) for (int j = 1; j <= num; j++) if (j + (1 << i) - 1 <= num) { mx[i][j] = max(mx[i - 1][j], mx[i - 1][j + (1 << i >> 1)]); mn[i][j] = min(mn[i - 1][j], mn[i - 1][j + (1 << i >> 1)]); }}int main (){ scanf(%d%d, &N, &Q); for (int i = 1; i <= N; i++) { scanf(%d, &mx[0][i]); mn[0][i] = mx[0][i]; } RMQ(N); int st, ed; while(Q--) { scanf(%d%d, &st, &ed); int k = (int) log2(ed - st + 1.0); int mxans = max(mx[k][st], mx[k][ed - (1 << k) + 1]); int mnans = min(mn[k][st], mn[k][ed - (1 << k) + 1]); printf(%d, mxans - mnans); } return 0;}