POJ 3264 Balanced Lineup | st表

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<cmath>
 5 #define N 50010
 6 using namespace std;
 7 int rmin[N][21],rmax[N][21],n,l,r,q;
 8 int read()
 9 {
10     int ret=0,neg=1;
11     char j=getchar();
12     for (;j>‘9‘ || j<‘0‘;j=getchar())
13     if (j==‘-‘) neg=-1;
14     for (;j<=‘9‘ && j>=‘0‘;j=getchar())
15     ret=ret*10+j-‘0‘;
16     return ret*neg;
17 }
18 int main()
19 {
20     n=read(),q=read();
21     for (int i=1,x;i<=n;i++)
22     x=read(),rmin[i][0]=x,rmax[i][0]=x;
23     for (int j=1;j<=20;j++)
24     for (int i=1;i+(1<<j-1)<=n;i++)
25         rmin[i][j]=min(rmin[i][j-1],rmin[i+(1<<j-1)][j-1]),rmax[i][j]=max(rmax[i][j-1],rmax[i+(1<<j-1)][j-1]);
26     while (q--)
27     {
28     l=read(),r=read();
29     int lg=log2(r-l+1);
30     printf("%d\n",max(rmax[l][lg],rmax[r-(1<<lg)+1][lg]) - min(rmin[l][lg],rmin[r-(1<<lg)+1][lg]));
31     }
32     return 0;
33 }