bzoj1699 [Usaco2007 Jan]Balanced Lineup排队

题目

区间最小最大值，还不带修改，随意做吧，用ST表水过。。。

#include<bits/stdc++.h>#define N 50000using namespace std;int n,q,l,r;int mx[N+5][25],mn[N+5][25];inline char nc(){    static char buf[100000],*p1=buf,*p2=buf;    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline int read(){    int x=0,b=1;    char c=nc();    for(;!(c<='9'&&c>='0');c=nc())if(c=='-')b=-1;    for(;c<='9'&&c>='0';c=nc())x=x*10+c-'0';    return x*b;}inline void write(int x){    if(x==0)putchar('0');    else    {        char buf[15];        int len=0;        if(x<0)putchar('-'),x=-x;        while(x)buf[++len]=x%10+'0',x/=10;        for(int i=len;i>=1;i--)putchar(buf[i]);    }    putchar('\n');}inline int get_mx(int l,int r){    int x=log2(r-l+1);    return max(mx[l][x],mx[r-(1<<x)+1][x]);}inline int get_mn(int l,int r){    int x=log2(r-l+1);    return min(mn[l][x],mn[r-(1<<x)+1][x]);}int main(){    freopen("in.txt","r",stdin);    n=read();q=read();    for(int i=1;i<=n;i++)mn[i][0]=mx[i][0]=read();    for(int i=1;i<=20;i++)        for(int j=1;j+(1<<i)-1<=n;j++)            mn[j][i]=min(mn[j][i-1],mn[j+(1<<(i-1))][i-1]),            mx[j][i]=max(mx[j][i-1],mx[j+(1<<(i-1))][i-1]);    for(int i=1;i<=q;i++)        l=read(),r=read(),write(get_mx(l,r)-get_mn(l,r));     return 0;}