1<=N<=10^18

鸣谢胡泽聪提供SPJ

 #include<iostream>

 #include<cstdio>

 #include<algorithm>

 #include<cstring>

 #define LL long long

 using namespace std;

 const int mxn=;

 LL n;

 LL b[];

 double p;

 int len;

 double pw(){//加密

     double res=;

     for(int i=;i<len;i++){

         double L=(n/(1LL<<(i+)))*(double)(1LL<<i);

         //高位取遍所有情况，当前位为0，低位无限制

         double R=(max(0LL,n%(1LL<<(i+))-(1LL<<i)));

         //高位为极限，当前位为1,低位不超限

         L=(L+R)/n;

         res+=L*(-L)**(1LL<<i);

         //取对立情况使得该位异或和为1的期望*贡献  x,y交换是另一种情况，所以*2

     }

     return res;

 }

 double f[];

 double npw(){//未加密

     n--;//右侧开区间

     double res=;

     if(n&)f[]=;else f[]=;

     f[]/=(n+);

     LL m=n;

     int i,j;

     for(i=;i<len-;i++){

         if((n>>i)&)f[i]=f[i-]+(double)b[i]/(n+)*b[i]*+(double)(b[i]-)*1.0/(n+)*b[i];

         else f[i]=f[i-]*+(b[i])/(double)(n+)*(b[i]);

     }

     for(int i=len-;i>=;i--){

         if((n>>i)&){//x可以取1

             if((m>>i)&){//y可以取1

                 res+=(b[i+]-)*(double)(m+-(b[i]))/(n+);

                 m=(b[i])-;

             }

             res+=b[i]*(double)(m+)/(n+);

         }

         else{

             if((m>>i)&){

                 res+=(b[i])*(double)(m+-(b[i]))/(n+);

                 m^=(b[i]);

                 res+=i->=?f[i-]:;

             }

         }

     }

     n++;

     return res;

 }

 int main(){

     int i,j;

     scanf("%lld%lf",&n,&p);

     len=;

     while((1LL<<len)<=n)++len;

     for(i=;i<=len;i++)b[i]=1LL<<i;

     printf("%.8f\n",npw()*p+pw()*(-p));

     return ;

 }