[Usaco07Jan&luogu2880] Balanced Lineup

题目链接

题解：rmq

#include <cstdio>#include <cmath>#include <iostream>using namespace std;#define pow2(x) (1<<(x))const int M=50005;int n,T,x,y;int a[M],Fmin[M][25],Fmax[M][25];void max_rmq(){    for(int i=1;i<=n;i++)    Fmax[i][0]=a[i];    for(int j=1;j<=20;j++)    for(int i=1;i+pow2(j)-1<=n;i++)    Fmax[i][j]=max(Fmax[i][j-1],Fmax[i+pow2(j-1)][j-1]);    }int max_query(int i,int j){    int k=log(j-i+1)/log(2);    return max(Fmax[i][k],Fmax[j-pow2(k)+1][k]);  }void min_rmq(){    for(int i=1;i<=n;i++)    Fmin[i][0]=a[i];    for(int j=1;j<=20;j++)    for(int i=1;i+pow2(j)-1<=n;i++)    Fmin[i][j]=min(Fmin[i][j-1],Fmin[i+pow2(j-1)][j-1]);    }int min_query(int i,int j){    int k=log(j-i+1)/log(2);    return min(Fmin[i][k],Fmin[j-pow2(k)+1][k]);  }int main(){    cin>>n>>T;    for(int i=1;i<=n;i++)    scanf("%d",&a[i]);    min_rmq();    max_rmq();    while(T--)    {        scanf("%d%d",&x,&y);        printf("%d\n",max_query(x,y)-min_query(x,y));    }    return 0;}