poj 3264 Balanced Lineup（查询区间最大值与最小值的差）

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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 100000int 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;}

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