poj Balanced Lineup（3264）！

此题也算技巧提吧！dp预处理将

#include <stdio.h>
#include <math.h>

int dpmax[50010][20],dpmin[50010][20],n,a[50010];
int max (int x,int y)
{
    return x>y?x:y;
}
int min (int x,int y)
{
    return x<y?x:y;
}
void init ()
{
    int i,j;
    for(i=1;i<=n;i++)
    {
       dpmax[i][0]=a[i];
       dpmin[i][0]=a[i];
    } 
    for(j=1;j<=log((double)(n+1))/log(2.0);j++)
    for(i=1;i+((1<<j)-1)<=n;i++)
    {
        dpmax[i][j]=max(dpmax[i][j-1],dpmax[i+(1<<j-1)][j-1]);
        dpmin[i][j]=min(dpmin[i][j-1],dpmin[i+(1<<j-1)][j-1]);
    } 
}
int getmax( int a , int b ){
    int k = (int)(log((double)(b-a+1))/log(2.0));
    return max( dpmax[a][k] , dpmax[b-(1<<k)+1][k] );    
}
int getmin( int a , int b ){
    int k = (int)(log((double)(b-a+1))/log(2.0));
    return min( dpmin[a][k] , dpmin[b-(1<<k)+1][k] );    
}
int main ()
{
    int m,i,j;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        init();
        while(m--)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            printf("%d
",getmax(a,b)-getmin(a,b));
        }
    }
}