【Codeforces441E】Valera and Number [DP]

Valera and Number

Time Limit: 20 Sec  Memory Limit: 512 MB

Description

　　

Input

　　

Output

　　

Sample Input

　　5 3 25

Sample Output

　　1.9218750000

HINT

　　

Solution

　　考虑运用DP。

　　

Code

 1 #include<iostream>  
 2 #include<string>  
 3 #include<algorithm>  
 4 #include<cstdio>  
 5 #include<cstring>  
 6 #include<cstdlib>  
 7 #include<cmath>
 8 using namespace std;
 9 typedef long long s64;
10  
11 const int ONE = 800005;
12 const int MOD = 1e9 + 7;
13 const int all = 255;
14 
15 int x, n;
16 double p;
17 double f[205][260][255][2];
18 double Ans;
19 int val, num;
20 
21 int get() 
22 { 
23         int res;char c; 
24         while( (c=getchar())<48 || c>57 );
25         res=c-48;  
26         while( (c=getchar())>=48 && c<=57 ) 
27         res=res*10+c-48; 
28         return res; 
29 } 
30 
31 void Deal_first(int n)
32 {
33         int record[250], num = 0, x = n;
34         while(x)
35             record[++num] = x & 1,
36             x >>= 1;
37         int j = 1, pos = 9, val = record[pos];
38         for(pos = 9; pos + j - 1 <= num; j++)
39             if(record[pos + j] != val) break;
40 
41         f[0][n & all][j][val] = 1; 
42 }
43 
44 int Get(int x)
45 {
46         int res = 0;
47         while(x % 2 == 0) res++, x >>= 1;
48         return res;
49 }
50 
51 int main()
52 {
53         x = get(), n = get(), p = (double)get() / 100;
54         Deal_first(x);
55 
56         for(int i = 0; i < n; i++)
57             for(int s = 0; s <= all; s++)
58                 for(int j = 1; j <= 250; j++)
59                     for(int k = 0; k <= 1; k++)
60                     {
61                         val = s >> 8 - 1 & 1;
62                         if(val != k) num = 1; else num = j + 1;
63                         f[i + 1][s << 1 & all][num][val] += f[i][s][j][k] * p;
64 
65                         val = s == all ? (k ^ 1) : k;
66                         if(val != k && k == 0) num = 1; else num = j;
67                         f[i + 1][s + 1 & all][num][val] += f[i][s][j][k] * (1 - p);
68                     }
69 
70         for(int s = 0; s <= all; s++)
71             for(int j = 1; j <= 250; j++)
72                 for(int k = 0; k <= 1; k++)
73                 {
74                     if(s == 0 && k == 0) val = j + 8;
75                     else if(s == 0 && k == 1) val = 8;
76                     else val = Get(s);
77                     Ans += f[n][s][j][k] * val;
78                 }
79 
80         printf("%.8lf", Ans);
81 }
View Code

 

  • Unkonwn
posted @ 2017-10-17 17:48  BearChild  阅读(175)  评论(0)  编辑  收藏  举报