hdu contest day1 1002 Assignment

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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;}

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