【Codeforces711E】ZS and The Birthday Paradox [数论]

ZS and The Birthday Paradox

Time Limit: 20 Sec  Memory Limit: 512 MB

Description

　　

Input

　　

Output

　　

Sample Input

　　4 3

Sample Output

　　23 128

HINT

　　

Solution

　　

Code

 1 #include<iostream>
 2 #include<string>
 3 #include<algorithm>
 4 #include<cstdio>
 5 #include<cstring>
 6 #include<cstdlib>
 7 #include<cmath>
 8 #include<vector>
 9 using namespace std;
10 typedef long long s64;
11 
12 const int ONE = 1000005;
13 const int MOD = 1e6 + 3;
14 
15 s64 n, k;
16 s64 num = 0;
17 s64 Fz = 1, Fm = 1;
18 
19 int get() 
20 { 
21         int res;char c; 
22         while( (c=getchar())<48 || c>57 );
23         res=c-48;  
24         while( (c=getchar())>=48 && c<=57 )
25         res=res*10+c-48; 
26         return res; 
27 }
28 
29 s64 Quickpow(s64 a, s64 b)
30 {
31         s64 res = 1;
32         while(b)
33         {
34             if(b & 1) res = res * a % MOD;
35             a = a * a % MOD;
36             b >>= 1;
37         }
38         return res;
39 }
40 
41 int main()
42 {
43         cin>>n>>k;
44 
45         num = 1;
46         while((1LL << num) < k) num++;
47         if(num > n) {printf("1 1"); return 0;}
48 
49         num = 0;
50         for(s64 i = k - 1; i >= 1; i >>= 1)
51             num += (i >> 1);
52 
53         s64 a = Quickpow(2, n);
54 
55         Fz = 1;
56         for(int i = 1; i < k; i++)
57         {
58             Fz = (s64)Fz * (a - i + MOD) % MOD;
59             if(Fz == 0) break;
60         }
61         Fm = Quickpow(a, k - 1);
62 
63         int inv = Quickpow(Quickpow(2, num), MOD - 2);
64 
65         Fz = (s64)Fz * inv % MOD, Fm = (s64)Fm * inv % MOD;
66         Fz = (Fm - Fz + MOD) % MOD;
67 
68 
69         cout<<Fz<<" "<<Fm<<endl;
70 }
View Code

 

posted @ 2017-10-18 17:20  BearChild  阅读(205)  评论(0)  编辑  收藏  举报