E. ZS and The Birthday Paradox
题目连接：
http://www.codeforces.com/contest/711/problem/E
Description
ZS the Coder has recently found an interesting concept called the Birthday Paradox. It states that given a random set of 23 people, there is around 50% chance that some two of them share the same birthday. ZS the Coder finds this very interesting, and decides to test this with the inhabitants of Udayland.
In Udayland, there are 2n days in a year. ZS the Coder wants to interview k people from Udayland, each of them has birthday in one of 2n days (each day with equal probability). He is interested in the probability of at least two of them have the birthday at the same day.
ZS the Coder knows that the answer can be written as an irreducible fraction . He wants to find the values of A and B (he does not like to deal with floating point numbers). Can you help him?
Input
The first and only line of the input contains two integers n and k (1 ≤ n ≤ 1018, 2 ≤ k ≤ 1018), meaning that there are 2n days in a year and that ZS the Coder wants to interview exactly k people.
Output
If the probability of at least two k people having the same birthday in 2n days long year equals (A ≥ 0, B ≥ 1, ), print the A and B in a single line.
Since these numbers may be too large, print them modulo 106 + 3. Note that A and B must be coprime before their remainders modulo 106 + 3 are taken.
Sample Input
3 2
Sample Output
1 8
Hint
题意
有(2^n)天，有(k)个小朋友，问你这些小朋友在这n天，至少有两个小朋友的生日在同一天的概率是多少，分子分母 mod 1e6+3
题解：
首先容斥，这个很简单。
最难的就是约分，然后我们考虑约分这个玩意儿，他肯定是除以gcd，显然gcd是2的幂，分母的幂显然比分子多，那么我统计一下分子有多少个2 就好了
如果k>=mod，显然答案为0，否则我就暴力。
然后就完了。
特判掉，人比天数多的情况
代码
#include<bits/stdc++.h>
using namespace std;
const int mod = 1e6+3;
long long quickpow(long long m,long long n,long long k)//返回m^n%k
{
long long b = 1;
while (n > 0)
{
if (n & 1)
b = (b*m)%k;
n = n >> 1 ;
m = (m*m)%k;
}
return b;
}
long long gcd(long long a,long long b)
{
if(b==0)return a;
return gcd(b,a%b);
}
int main()
{
long long n,k;
cin>>n>>k;
if(n<62&&k>(1LL<<n))return puts("1 1"),0;
long long num = n;
for(long long i=1;i<62;i++)
num+=(k-1)/(1LL<<i);
long long A=1;
if(k<mod)
{
for(long long i=1;i<=k;i++)A=A*(quickpow(2,n,mod)-i+mod+1)%mod;
A=A*quickpow(quickpow(2,mod-2,mod),num,mod)%mod;
}
else
A=0;
long long B = quickpow(quickpow(2,n,mod),k,mod)*quickpow(quickpow(2,mod-2,mod),num,mod)%mod;
cout<<(B-A+mod)%mod<<" "<<B<<endl;
}