Submit: 215 Solved: 112
Description
Input
Output
Sample Input
3 0.5
Sample Output
2.000000
HINT
1<=N<=10^18
Source
加密和不加密各自是独立问题
后者是炒鸡麻烦的数位DP
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cstring> 5 #define LL long long 6 using namespace std; 7 const int mxn=100010; 8 LL n; 9 LL b[70]; 10 double p; 11 int len; 12 double pw(){//加密 13 double res=0; 14 for(int i=0;i<len;i++){ 15 double L=(n/(1LL<<(i+1)))*(double)(1LL<<i); 16 //高位取遍所有情况，当前位为0，低位无限制 17 double R=(max(0LL,n%(1LL<<(i+1))-(1LL<<i))); 18 //高位为极限，当前位为1,低位不超限 19 L=(L+R)/n; 20 res+=L*(1-L)*2*(1LL<<i); 21 //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 22 } 23 return res; 24 } 25 double f[200]; 26 double npw(){//未加密 27 n--;//右侧开区间 28 double res=0; 29 if(n&1)f[0]=2;else f[0]=1; 30 f[0]/=(n+1); 31 LL m=n; 32 int i,j; 33 for(i=1;i<len-1;i++){ 34 if((n>>i)&1)f[i]=f[i-1]+(double)b[i]/(n+1)*b[i]*2+(double)(b[i]-1)*1.0/(n+1)*b[i]; 35 else f[i]=f[i-1]*2+(b[i])/(double)(n+1)*(b[i]); 36 } 37 for(int i=len-1;i>=0;i--){ 38 if((n>>i)&1){//x可以取1 39 if((m>>i)&1){//y可以取1 40 res+=(b[i+1]-1)*(double)(m+1-(b[i]))/(n+1); 41 m=(b[i])-1; 42 } 43 res+=b[i]*(double)(m+1)/(n+1); 44 } 45 else{ 46 if((m>>i)&1){ 47 res+=(b[i])*(double)(m+1-(b[i]))/(n+1); 48 m^=(b[i]); 49 res+=i-1>=0?f[i-1]:0; 50 } 51 } 52 } 53 n++; 54 return res; 55 } 56 int main(){ 57 int i,j; 58 scanf("%lld%lf",&n,&p); 59 len=1; 60 while((1LL<<len)<=n)++len; 61 for(i=0;i<=len;i++)b[i]=1LL<<i; 62 printf("%.8f ",npw()*p+pw()*(1-p)); 63 return 0; 64 }