[交互 杂题] Codeforces Gym 100307 NEERC 13 I. Interactive Interception

如果已知起点 我们可以直接二分速度
现在我们都不知道
那么我们每次查询希望把起点和速度组成的二元组尽量平均的分开
这个要一个二分来找询问那个点
然后询问(0,R)就好了

#include<cstdio>#include<cstdlib>#include<algorithm>#include<string>#include<iostream>using namespace std;typedef long long ll;const int N=100005;int V,P;int l[N],r[N],mid[N];inline ll f(int t,int c){  ll ret=0;  for (int i=0;i<=P;i++){    if (l[i]>r[i]) continue;    int mv=c-i;    if (!t)      mv=mv>=0?V:-1;    else      mv=mv<0?-1:mv/t;    mid[i]=mv;    ret+=mv>=r[i]?r[i]-l[i]+1:mv-l[i]+1;  }  return ret;}int main(){  string ret;  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  cin>>P>>V;  for (int i=0;i<=P;i++) l[i]=0,r[i]=V;  ll tot=(ll)(V+1)*(P+1);  for (int t=0;;t++){    int L=-1,R=P+t*V,MID;    while (L+1<R)      if (f(t,MID=(L+R)>>1)*2>=tot)    R=MID;      else    L=MID;    f(t,R);    printf("check 0 %d\n",R); fflush(stdout);    cin>>ret;    if (ret=="Yes"){      for (int i=0;i<=P;i++) if (mid[i]<r[i]) r[i]=mid[i];    }else{      for (int i=0;i<=P;i++) if (mid[i]+1>l[i]) l[i]=mid[i]+1;    }    tot=0;    for (int i=0;i<=P;i++) if (l[i]<=r[i]) tot+=r[i]-l[i]+1;    if (tot<=1)      for (int i=0;i<=P;i++)        if (l[i]<=r[i]){      printf("answer %d\n",i+(t+1)*l[i]); fflush(stdout);      exit(0);    }  }  return 0;}