time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.
Examples
input
3 2
output
1 8
input
1 3
output
1 1
input
4 3
output
23 128
Note
In the first sample case, there are 2^3 = 8 days in Udayland. The probability that 2 people have the same birthday among 2 people is clearly 1/8, so A = 1, B = 8.
In the second sample case, there are only 2^1 = 2 days in Udayland, but there are 3 people, so it is guaranteed that two of them have the same birthday. Thus, the probability is 1 and A = B = 1.
【题解】
每个人的生日都有2^n个可能;然后有k个人;
要求的是k个人里面至少有两个人的生日相同的概率;
正难则反；
求出所有人的生日都不同的概率p;
再用1减去p就可以了;
p=A(2^n,k)/((2^n)^k);
A是排列数;
写成乘法的形式就变成
p=(2^n)(2^n-1)(2^n-2)···(2^n-(k-1))/((2^n)^k)
分子分母同时除2^n;
(2^n-1)(2^n-2)···*(2^n-(k-1))/((2^(k-1))^n);
然后就是约分了;
分子的2的数目比分母少;
那么公约数就是2^temp次方了;
temp是分子的所含的因子2的个数;
而(2^n-1)···(2^n-(k-1))中
2^n-i的2因子的个数显然是i决定的,i有几个2因子,这项就有几个2因子;
比如2^n-2,2有一个2因子，所以这项有1个2因子;
实际上就可以转化为(k-1)!的2因子的个数;
有个关于阶乘(k-1)!素因子p的公式;
temp = ∑(k-1)/i;
其中i = p^1、p^2、p^3..p^m;
且2^m<= k-1;
求出来个数就好;
设为temp;
则可以约掉的数就是2^temp
分子和分母都要除2^temp；
但是要求余?
除法求余?
求乘法逆元!
乘法逆元?
比如要求(a/b)%p;
且(b*k)%p==1;
则(a/b)%p == (a*k)%p;
这个k就是b的乘法逆元。(可能有定义不对的地方。谅解下);
同时a/b一定要为整数;
证明:
因为(b*k)%p=1
所以b*k = p*x+1;
k = (p*x+1)/b;
则(a*k)%p=(apx/b+a/b)%p = ((a/b)*x*p)%p+(a/b)%p;
因为b能够整除a,所以a/b为整数,又乘上了p，则%p不就为0吗;
则(a*k)%p == (a/b)%p;
如何求这个k
b*k = p*x+1;
->k*b+(-x)*p=1
;
即解一个二元一次方程组;
->用扩展欧几里得算法求解;
扩展欧几里得算法?
ax+by=gcd(a,b);
这里如果a和b互质(因为p是质数而b是肯定小于p的(因为要取余嘛),所以b和p肯定是互质的);
ax+by=1
这里进行一下递推;
设
x1a+y1b=gcd(a,b);
x2b+y2(a%b) = gcd(b,a%b);
而又欧几里得算法gcd(a,b)==gcd(b,a%b);
所以x1a+y1b=x2b+y2(a%b);
a%b可以写出a-(a/b)*b 这里的/是整除
则
x1a+y1b=x2b+y2(a-(a/b)*b)
x1a+y1b=x2b+y2a-y2(a/b)*b
x1a+y1b=y2a+(x2-(a/b)*y2)*b
->x1=y2
->y1 =x2-(a/b)*y2
根据这个递推式
可以写出扩展欧几里得算法的程序
void ex_gcd(LL a,LL b,LL &x,LL &y)
{
if (b == 0)//gcd(a,b)==gcd(a,0)==a;所以要使得xa+yb==gcd(a,b)只要让x==1,y==0即可
{
x = 1;y = 0;
return;
}
ex_gcd(b,a % b,x,y);
LL temp = y;
y = x-(a/b)*temp;
x = temp;
}
我们只要执行
ex_gcd(2^temp,p,ni,useless);
我们要的是这个方程的k
k*b+(-x)*p=1
所以最后得到的ni就是k，也即2^temp的乘法逆元;
对于分母直接乘上这个ni。就表示除去了公约数;
对于分子
(2^n-1)(2^n-2)···*(2^n-(k-1))
如果k-1>=mod;则我们最少得到了连续的mod个数;
则这里面肯定有mod的倍数;
所以此时分子为0;
直接输出 分母-‘0’ 分母即可;
对于k-1小于mod的情况，这个时候k很小了。直接暴力求解
(2^n-1)(2^n-2)···*(2^n-(k-1))%mod即可;
然后输出 (分母-分子+mod)%mod 分母 即可;
#include <cstdio>
#include <algorithm>
#define LL long long
using namespace std;
const LL mod = 1e6+3;
const int INF = 63;
LL n,k,tmp = 0,ni,fz,fm;
LL ksm(LL x,LL y)
{
if (y == 0)
return 1;
LL temp =ksm(x,y>>1);
temp = (temp*temp)%mod;
if (y&1)
temp = (temp*x)%mod;
return temp;
}
void ex_gcd(LL a,LL b,LL &x,LL &y)
{
if (b == 0)
{
x = 1;y = 0;
return;
}
ex_gcd(b,a % b,x,y);
LL temp = y;
y = x-(a/b)*temp;
x = temp;
}
int main()
{
scanf("%I64d%I64d",&n,&k);
bool flag = false;
if (n >= 63)
flag = true;
else
{
LL temp = 1;
for (int i = 1;i <= n;i++)
{
temp = temp *2;
if (temp >=k)
{
flag = true;
break;
}
}
}
if (!flag)
{
puts("1 1");
return 0;
}
LL i;
for (i = 2;i <= (k-1);i<<=1)
tmp+=(k-1)/i;
tmp = ksm(2,tmp);
LL fm = ksm(ksm(2,k-1),n);
LL nu;//这个nu变量是没用的
ex_gcd(tmp,mod,ni,nu);
ni = (ni + mod) %mod;//求出来的ni是乘法逆元
fm = (fm * ni)%mod;
if (k-1>= mod)
printf("%I64d %I64d
",fm,fm);
else//暴力求解分子
{
LL a = ksm(2,n);
LL fz = 1;
for (i = 1;i <= k-1;i++)
fz = (fz*((a-i+mod) % mod))%mod;
fz=(fz*ni)%mod;
fz = (fm-fz+mod)%mod;
printf("%I64d %I64d
",fz,fm);
}
return 0;
}