联赛模拟17_简单的玄学

m个数各不相同概率为 \(\frac{A_{2^n}^{m}}{(2^n)^m}\)

答案就是
\(1-\frac{(2^n)*(2^n-1)...(2^n-m+1)}{(2^n)^m}\)

=\(\frac{(2^n)^m-(2^n)*(2^n-1)...(2^n-m+1)}{(2^n)^m}\)

分子分母先约2,再取模

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define int long long
const int M=1e6+3;
const int phi=1e6+2;
int n,m;
int erncf;
int fz,fm;
int cnt;
int qp(int a,int x){
	int ans=1;
	while(x){
		if(x&1) ans=ans*a%M;
		x>>=1,a=a*a%M;
	}
	return ans;
}
signed main(){
	freopen("random.in","r",stdin);
	freopen("random.out","w",stdout);
	scanf("%lld%lld",&n,&m);
	erncf=qp(2,n%phi);// 2的n次方
	fm=qp(erncf,m%phi);
	fz=1;
	for(int i=0;i<m&&fz;++i) fz=fz*(erncf-i)%M;
	cnt=n;
	for(int i=1;(1ll<<i)<=m-1;++i) cnt+=(m-1)/(1ll<<i);
	fz=fz*qp(qp(2,M-2),cnt)%M;
	fm=fm*qp(qp(2,M-2),cnt)%M;
	printf("%lld %lld\n",(fm-fz+M)%M,fm);
	return 0;
}