CodeForces 711 E.ZS and The Birthday Paradox（组合数学）

Description
一年2^n天，k个人，问这k个人中至少有两个人生日是同一天的概率
Input
两个整数n和k(1<=n<=1e18,2<=k<=1e18)
Output
将答案表示为最简分数形式A/B，输出A B，结果过大把A和B模1e6+3后输出
Sample Input
3 2
Sample Output
1 8
Solution

Code

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#include<cmath>#include<vector>#include<queue>#include<map>#include<set>#include<ctime>using namespace std;typedef long long ll;#define INF 0x3f3f3f3f#define maxn 1111const ll mod=1000003;ll mod_pow(ll a,ll b,ll c){    ll ans=1;    while(b)    {        if(b&1)ans=ans*a%c;        a=a*a%c;        b>>=1;    }    return ans;}ll Count(ll k){    ll ans=0;    while(k)ans+=k/2,k/=2;    return ans;}int main(){    ll n,k;    while(~scanf("%I64d%I64d",&n,&k))    {        int kk=0;        while((1ll<<kk)<k)kk++;        if(kk>n)        {            printf("1 1\n");            continue;        }        ll num=Count(k-1),a=mod_pow(2,n%(mod-1),mod),b=mod_pow(2,mod-1-num%(mod-1),mod);        ll ans1=mod_pow(a,k-1,mod)*b%mod;        if(k-1>=mod)printf("%I64d %I64d\n",ans1,ans1);        else        {            ll ans2=ans1,temp=1;            for(int i=1;i<k;i++)temp=temp*(a-i+mod)%mod;            temp=temp*b%mod;            ans2=(ans2-temp+mod)%mod;            printf("%I64d %I64d\n",ans2,ans1);        }    }    return 0;}