Bzoj3652 大新闻

Time Limit: 10 Sec  Memory Limit: 512 MBSec  Special Judge
Submit: 215  Solved: 112

Description

Input

Output

Sample Input

3 0.5

Sample Output

2.000000

HINT

1<=N<=10^18

Source

鸣谢胡泽聪提供SPJ

加密和不加密各自是独立问题

后者是炒鸡麻烦的数位DP

 #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 ;
 }