hdu contest day1 1002 Assignment

传送门：http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1002&cid=589

思路：对于每个人，向右二分判断能组成group最远的人，用ST表维护最大最小值，判断时只要看最大最小只差是否小于k即可

CYY的代码

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long int64;
const double eps=1e-7;
const int maxn=100015,maxk=20;
int n,m,k,a[maxn];int64 ans;
int bin[maxk],fmx[maxk][maxn],fmn[maxk][maxn];
void read(int &x){
    char ch;
    for (ch=getchar();!isdigit(ch);ch=getchar());
    for (x=0;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
}
void init(){
    read(n);read(m);
    for (int i=1;i<=n;++i) read(a[i]);
}
void prepare(){
    k=floor(log2(n)+eps);
    for (int i=1;i<=n;++i) fmx[0][i]=fmn[0][i]=a[i];
    for (int i=1;i<=k;++i)
        for (int j=bin[i];j<=n;++j){
            fmx[i][j]=max(fmx[i-1][j],fmx[i-1][j-bin[i-1]]);
            fmn[i][j]=min(fmn[i-1][j],fmn[i-1][j-bin[i-1]]);
        }
}
int qmax(int l,int r){
    int q=floor(log2(r-l+1)+eps);
    return max(fmx[q][r],fmx[q][l+bin[q]-1]);
}
int qmin(int l,int r){
    int q=floor(log2(r-l+1)+eps);
    return min(fmn[q][r],fmn[q][l+bin[q]-1]);
}
int query(int a,int b){
    int l=a,r=b,res;
    while (l<=r){
        int mid=(l+r)>>1;
        if (qmax(mid,b)-qmin(mid,b)<m) r=(res=mid)-1;
        else l=mid+1;
    }
    return res;
}
void work(){
    prepare();
    ans=0;for (int i=1;i<=n;++i) ans+=(i-query(1,i)+1);
    printf("%I64d\n",ans);
}
int main(){
    int cases;scanf("%d",&cases);
    bin[0]=1;for (int i=1;i<maxk;++i) bin[i]=bin[i-1]<<1;
    while (cases--){init();work();}
    return 0;
}