poj 3264 Balanced Lineup(查询区间最大值与最小值的差)
最后更新于:2022-04-01 16:02:10
1.代码:
~~~
#include<stdio.h>
#include<string.h>
#include<math.h>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define N 100000
int a[N];
int ST1[N][20];
int ST2[N][20];
int n,q;
void make_ST()
{
for(int j=1;(1<<j)<=n;j++)
{
for(int i=1;(i+(1<<j)-1)<=n;i++)
{
ST1[i][j]=Max(ST1[i][j-1],ST1[i+(1<<(j-1))][j-1]);
ST2[i][j]=Min(ST2[i][j-1],ST2[i+(1<<(j-1))][j-1]);
}
}
}
int Query(int l,int r)
{
int k=floor(log((double)(r-l+1))/log((double)(2)));
return Max(ST1[l][k],ST1[r-(1<<k)+1][k])-Min(ST2[l][k],ST2[r-(1<<k)+1][k]);
}
int main()
{
while(scanf("%d%d",&n,&q)==2)
{
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
ST1[i][0]=ST2[i][0]=a[i];
}
make_ST();
while(q--)
{
int l,r;
scanf("%d%d",&l,&r);
printf("%d\n",Query(l,r));
}
}
return 0;
}
~~~